Thinking Process

In this problem, we have to locate the graph which is having same displacement for two timings. When there are two timings for same displacement the corresponding velocities will be in opposite directions.

Answer (b) In graph (b) for one value of displacement there are two different points of time. Hence, for one time, the average velocity is positive and for other time is equivalent negative.

As there are opposite velocities in the interval 0 to $T$ hence average velocity can vanish in (b). This can be seen in the figure given below

Here, $OA=BT$ (same displacement) for two different points of time.

• Graph (a): In graph (a), the displacement is always increasing or decreasing monotonically over time. This means that the velocity does not change sign, and thus the average velocity over any interval cannot be zero.

• Graph (c): In graph (c), the displacement is a straight line with a constant slope, indicating a constant velocity. Since the velocity does not change, the average velocity over any interval will be equal to this constant velocity and cannot be zero.

• Graph (d): In graph (d), the displacement is also a straight line with a constant slope, similar to graph (c). This indicates a constant velocity, and thus the average velocity over any interval will be equal to this constant velocity and cannot be zero.

Answer (a) As the lift is coming in downward directions displacement will be negative. We have to see whether the motion is accelerating or retarding

We know that due to downward motion displacement will be negative. When the lift reaches 4th floor is about to stop hence, motion is retarding in nature hence, $x<0;a>0$.

As displacement is in negative direction, velocity will also be negative i.e., $v<0$.

This can be shown on the adjacent graph.

• Option (b): $x>0,v<0,a<0$

  • Reason: The displacement $x$ should be negative because the lift is moving downward from the 8th floor to the 4th floor. Therefore, $x>0$ is incorrect. Additionally, if the lift is about to stop, it is decelerating, which means the acceleration $a$ should be positive, not negative.

• Option (c): $x>0,v<0,a>0$

  • Reason: The displacement $x$ should be negative because the lift is moving downward from the 8th floor to the 4th floor. Therefore, $x>0$ is incorrect.

• Option (d): $x>0,v>0,a<0$

  • Reason: The displacement $x$ should be negative because the lift is moving downward from the 8th floor to the 4th floor. Therefore, $x>0$ is incorrect. Additionally, the velocity $v$ should be negative because the lift is moving downward, not upward.

Answer (b) For maximum and minimum displacement we have to keep in mind the magnitude and direction of maximum velocity.

As maximum velocity in positive direction is $v_0$ maximum velocity in opposite direction is also $v_0$.

Maximum displacement in one direction $=v_{0}T$

Maximum displacement in opposite directions $=-v_{0}T$

Hence, $-v_{0}T<x<v_{0}T$

Note We should not confuse with direction of velocities i.e., in one direction it is taken as positive and in another direction is taken as negative.

• (a) The displacement in time ( T ) can take negative values if the motion is in the opposite direction. Therefore, it is not always non-negative.

• (c) The acceleration can be negative if the object is decelerating. Hence, it is not always a non-negative number.

• (d) The motion can have turning points if the object changes direction. Therefore, the motion does not necessarily have no turning points.

Thinking Process

To calculate average speed we will calculate total distance covered and will divide by total time taken.

Answer (c) Time taken to travel first half distance $t_1=\frac{l/2}{v_1}=\frac{l}{2v_1}$

Time taken to travel second half distance $t_2=\frac{l}{2v_2}$

Total time $=t_1+t_2$

$=\frac{l}{2v_1}+\frac{l}{2v_2}=\frac{l}{2}[\frac{1}{v_1}+\frac{1}{v_2}]$

We know that,

$\begin{array}{rl}{V}_{av}=& \text{ Average speed }=\frac{\text{ total distance }}{\text{ total time }}\\ =& \frac{l}{\frac{l}{2}[\frac{1}{v_1}+\frac{1}{v_2}]}=\frac{2v_1{v}_2}{v_1+v_2}\end{array}$

Note We should not confuse with distance and displacement. Distance $\ge$ displacement.

• Option (a) $\frac{v_1+v_2}{2}$: This option represents the arithmetic mean of the two speeds. However, the average speed for a journey where different segments are traveled at different speeds is not simply the arithmetic mean of those speeds. The average speed must take into account the time spent traveling at each speed, which is not done in this option.

• Option (b) $\frac{2v_1+v_2}{v_1+v_2}$: This option does not correctly account for the relationship between distance, speed, and time. The formula for average speed must consider the harmonic mean of the speeds when the distance is divided equally, which this option does not.

• Option (d) $\frac{L(v_1+v_2)}{v_1{v}_2}$: This option incorrectly uses the total distance $L$ in the numerator and does not correctly represent the harmonic mean of the speeds. The correct formula for average speed when traveling equal distances at different speeds involves the harmonic mean, not this expression.

Thinking Process

In such type of problems we have to see whether the motion is accelerating or retarding. During retarding journey particle will stop in between.

Answer (b) Given,

$x=(t-2{)}^2$

Velocity,

$v=\frac{dx}{dt}=\frac{d}{dt}(t-2{)}^2=2(t-2)m/s$

Acceleration,

$a=\frac{dv}{dt}=\frac{d}{dt}[2(t-2)]$

$=2[1-0]=2m/s^2$

When,

$\begin{array}{rl}& t=0;v=-4m/s\\ & t=2s;v=0m/s\\ & t=4s;v=4m/s\end{array}$

$v$ - $t$ graph is shown in adjacent diagram.

Distance travelled $=$ area of the graph

$\begin{array}{rl}& =\text{ area }OAC+\text{ area }ABD\\ & =\frac{4\times 2}{2}+\frac{1}{2}\times 2\times 4=8m\end{array}$

• Option (a) $4m$: This option is incorrect because the calculation of the distance covered by the particle in the first 4 seconds, based on the given displacement function and the velocity-time graph, shows that the total distance is 8 meters. The area under the velocity-time graph, which represents the distance, is calculated as 8 meters, not 4 meters.

• Option (c) $12m$: This option is incorrect because the total distance covered by the particle, as derived from the velocity-time graph, is 8 meters. The calculation involves summing the areas of two triangles formed in the velocity-time graph, which results in 8 meters, not 12 meters.

• Option (d) $16m$: This option is incorrect because the total distance covered by the particle in the first 4 seconds is 8 meters, as determined by the area under the velocity-time graph. The correct calculation of the areas of the triangles in the graph gives a total distance of 8 meters, not 16 meters.

Answer (c) In this question, we have to find net velocity with respect to the Earth that will be equal to velocity of the girl plus velocity of escalator.

Let displacement is $L$,then

$\text{ velocity of girl }{v}_g=\frac{L}{t_1}$

velocity of escalator $v_e=\frac{L}{t_2}$

Net velocity of the girl $=v_g+v_e=\frac{L}{t_1}+\frac{L}{t_2}$

If $t$ is total time taken in covering distance $L$, then

$\frac{L}{t}=\frac{L}{t_1}+\frac{L}{t_2}\Rightarrow t=\frac{t_1{t}_2}{t_1+t_2}$

• Option (a): $(t_1+t_2)/2$ is incorrect because it represents the arithmetic mean of the two times, which does not account for the combined effect of the girl’s walking speed and the escalator’s speed. The correct formula involves the harmonic mean, not the arithmetic mean.

• Option (b): $t_1{t}_2/(t_2-t_1)$ is incorrect because it suggests a relationship where the time taken is inversely proportional to the difference between the two times. This does not correctly represent the combined effect of the girl’s walking speed and the escalator’s speed. The correct relationship involves the sum of the times, not the difference.

• Option (d): $t_1-t_2$ is incorrect because it implies that the time taken is simply the difference between the two times, which does not take into account the combined velocities of the girl and the escalator. The correct formula involves the harmonic mean of the two times, not their difference.

Answer $(a,c,d)$

When we are calculating velocity of a displacement-time graph we have to take slope similarly we have to take slope of velocity-time graph to calculate acceleration. When slope is constant motion will be uniform.

When we are representing motion by a graph it may be displacement-time, velocity-time or acceleration-time hence, $B$ may represent time. For uniform motion velocity-time graph should be a straight line parallel to time axis. For uniform motion velocity is constant hence, slope will be positive. Hence quantity $A$ is displacement.

For uniformly accelerated motion slope will be positive and $A$ will represent velocity.

• Option (b) is incorrect: Quantity A cannot be velocity if the motion is uniform because, in uniform motion, the velocity is constant and does not vary with time. The graph would show a horizontal line if A were velocity and B were time, which is not depicted in the given graph.

Thinking Process

In this problem, we have to apply formula for slope $=\frac{dx}{dt}$ for the graph and velocity

$v=\frac{dx}{dt}$.

Answer $(a,c,d)$

As per the diagram, at point $A$ the graph is parallel to time axis hence, $v=\frac{dx}{dt}=0$. As the starting point is $A$ hence, we can say that the particle is starting from rest.

At $C$, the graph changes slope, hence, velocity also changes. As graph at $C$ is almost parallel to time axis hence, we can say that velocity vanishes.

As direction of acceleration changes hence, we can say that it may be zero in between.

From the graph it is clear that

$\mid \text{ slope at }D|>|\text{ slope at }E\mid$

Hence, speed at $D$ will be more than at $E$.

Note We should be very clear about magnitude of slope. Negative slope does not mean less value. It represents change in direction of velocity.

• Option (b): At point B, the graph is concave downwards, indicating that the slope of the graph (velocity) is decreasing. This implies that the acceleration is negative, not positive. Therefore, the statement that the acceleration $a>0$ at B is incorrect.

Answer $(a,d)$

Given,

$x=t-\sin t$

$\text{ velocity }v=\frac{dx}{dt}=\frac{d}{dt}[t-\sin t]$

=$1-\cos t$

$\text{ Acceleration }a=\frac{dv}{dt}=\frac{d}{dt}[1-\cos t]=\sin t$

$a>0\text{ for all }t>0$

$x(t)>0\text{ for all }t>0$

$\text{ Velocity }v=1-\cos t$

$\cos t=1,\text{ velocity }v=0$

$v_{max}=1-(\cos t{)}_{min}=1-(-1)=2$

$v_{min}=1-(\cos t{)}_{max}=1-1=0$

As acceleration

Hence,

When,

Hence, $v$ lies between 0 and 2

$\text{ Acceleration }a=\frac{dv}{dt}=-\sin t$

When $t=0;x=0,x=+1,a=0$

When $t=\frac{\pi }{2};x=1,v=0,a=-1$

When $t=\pi ;x=0,x=-1,a=1$

When $t=2\pi ;x=0,x=0,a=0$

Note (i) When sinusoidal function is involved in an expression we should be careful about sine and cosine functions.

(ii) We should be very careful when calculating maximum and minimum value of velocity because it involves inverse relation with cost in the given expression.

• Option (b) is incorrect: The velocity ( v(t) = 1 - \cos t ) is not greater than 0 for all ( t > 0 ). For example, when ( t = 2n\pi ) (where ( n ) is an integer), ( \cos t = 1 ) and thus ( v(t) = 0 ). Therefore, ( v(t) ) is not always positive.

• Option (c) is incorrect: The acceleration ( a(t) = \sin t ) is not greater than 0 for all ( t > 0 ). The sine function oscillates between -1 and 1, so there are intervals where ( \sin t ) is negative, making ( a(t) ) negative during those intervals. Therefore, ( a(t) ) is not always positive.

Answer (a, c)

When spring is stretched by $x$, restoring force will be $F=-kx$

Potential energy of the stretched spring $=PE=\frac{1}{2}k{x}^2$

The restoring force is central hence, when particle released it will execute SHM about equilibrium position.

Acceleration will be

$a=\frac{F}{m}=\frac{-kx}{m}$

$\begin{array}{cc}\text{ At equilibrium position, }& x=0\Rightarrow a=0\\ \text{ Hence, when just released }& x=x_{max}\end{array}$

Hence, acceleration is maximum. Thus option (a) is correct.

At equilibrium whole $PE$ will be converted to $KE$ hence, $KE$ will be maximum and hence, speed will be maximum.

• (b) Magnitude of acceleration, when at equilibrium position, is maximum: This is incorrect because at the equilibrium position, the displacement ( x ) is zero. Since acceleration ( a ) is given by ( a = \frac{-kx}{m} ), when ( x = 0 ), ( a ) will also be zero. Therefore, the magnitude of acceleration is not maximum at the equilibrium position; it is actually zero.

• (d) Magnitude of displacement is always maximum whenever speed is minimum: This is incorrect because the magnitude of displacement is maximum at the extreme positions of the oscillation, where the speed is indeed zero. However, the statement “always” implies that whenever the speed is minimum, the displacement is maximum, which is not true. The speed is also zero at the turning points, but it is not necessarily minimum at all other points where the displacement is not maximum. Therefore, the statement is not universally true.

Answer $(b,c,d)$

In this problem, we have to keep in mind the frame of the observer. Here we must be clear that we are considering the motion from the ground. Compare to velocity of trains $(10m/s)$ speed of ball is less $(1m/s)$.

The speed of the ball before collision with side of train is $10+1=11m/s$.

Speed after collision with side of train $=10-1=9m/s$.

As speed is changing after travelling $10m$ and speed is $1m/s$ hence, time duration of the changing speed is 10 .

Since, the collision of the ball is perfectly elastic there is no dissipation of energy hence, total momentum and kinetic energy are conserved.

Since, the train is moving with constant velocity hence, it will act as inertial frame of reference as that of Earth and acceleration will be same in both frames.

We should not confuse with non-inertial and inertial frame of reference. A frame of reference that is not accelerating will be inertial.

• (a) The direction of motion of the ball changes every 10: This option is incorrect because the direction of motion of the ball changes every time it collides with the walls of the railway compartment. Given that the ball is bouncing elastically between the walls and the compartment is 10 meters wide, the ball will change direction every 10 seconds in the frame of reference of the train. However, from the ground frame of reference, the ball’s speed changes due to the relative motion of the train, but the direction change is not strictly every 10 seconds due to the combined motion of the train and the ball.

Answer We have to analyse slope of each curve i.e., $\frac{dx}{dt}$. For peak points $\frac{dx}{dt}$ will be zero as $x$ is maximum at peak points.

For graph (a), there is a point (B) for which displacement is zero. So, a matches with (iii) In graph (b), $x$ is positive ( $>0$ ) throughout and at point $B_1,V=\frac{dx}{dt}=0$. since, at point of curvature changes $a=0$, So $b$ matches with (ii) In graph (c), slope $V=\frac{dx}{dt}$ is negative hence, velocity will be negative. so matches wth (iv) In graph (d), as slope $V=\frac{dx}{dt}$ is positive hence, $V>0$ Hence, $d$ matches with (i)

Answer If gravity effect is neglected then ball moving uniformly turned back with same speed when a bat hit it. Acceleration of the ball is zero just before it strikes the bat. When the ball strikes the bat, it gets accelerated due to the applied impulsive force by the bat.

The variation of acceleration with time is shown in graph

Answer When we are writing an equation belonging to periodic nature it will involve sine or cosine function.

(a) The particle will be moving along positive $x$-direction only if $t>\sin t$

Hence,

$\begin{array}{rl}x(t)& =1-\sin t\\ \text{ Velocity }v(t)& =\frac{dx(1)}{dt}=1-\cos t\\ \text{ Acceleration }a(t)& =\frac{dv}{dt}=\sin t\end{array}$

When $t=0;x(t)=0$

When $t=\pi ;x(t)=\pi >0$

When $t=0;x(t)=2\pi >0$

(b) Equation can be represented by

$\begin{array}{rl}x(t)& =\sin t\\ v& =\frac{d}{dt}x(t)=\cos t\end{array}$

As displacement and velocity is involving sint and cost hence these equations represent periodic.

Answer Let the motion is represented by

Suppose we are considering any instant $t$, then from Eq. (i), we can say that

$x(t)>0;v(t)<0\text{ and }a>0$

Answer When speed becomes constant acceleration $a=\frac{dv}{dt}=0$

Given acceleration

$a=g-bv$

where, $g$ = gravitational acceleration

Clearly, from above equation as speed increases acceleration will decrease. At a certain speed say $v_0$, acceleration will be zero and speed will remain constant.

Hence,

$a=g-b{v}_0=0$

$\Rightarrow$

$v_0=g/b$

Thinking Process

To calculate velocity we will find slope $\frac{dx}{dt}$ for displacement-time curve and to find

acceleration we will find slope $\frac{dV}{dt}$ of velocity-time curve.

Answer It is clear from the graph that displacement $x$ is positive throughout. Ball is dropped from a height and its velocity increases in downward direction due to gravity pull. In this condition $v$ is negative but acceleration of the ball is equal to acceleration due to gravity i.e., $a=-g$. When ball rebounds in upward direction its velocity is positive but acceleration is $a=-g$.

(a) The velocity-time graph of the ball is shown in fig. (i).

(i)

(b) The acceleration-time graph of the ball is shown in fig. (ii).

Thinking Process

First we have to calculate velocity and acceleration and then we can determine the maximum or minimum value accordingly.

Answer Given,

$\begin{array}{rl}& x(t)=x_0(1-e^{-\gamma t})\\ & v(t)=\frac{dx(t)}{dt}=x_0\gamma e^{-\gamma t}\\ & a(t)=\frac{dv(t)}{dt}=-x_0{\gamma }^2{e}^{-\gamma t}\end{array}$

(a) Whent $=0;x(t)=x_0(1-e^{-0})=x_0(1-1)=0$

$x(t=0)=x_0\gamma e^{-0}=x_0\gamma (1)=\gamma x_0$

(b) $x(t)$ is maximum when $t=\mathrm{\infty }$

$[x(t){]}_{max}=x_0$

$x(t)$ is minimum when $t=0[x(t){]}_{\text{min }}=0$

$v(t)$ is maximum when $t=0;v(0)=x_0\gamma$

$v(t)$ is minimum when $t=\mathrm{\infty };v(\mathrm{\infty })=0$

$a(t)$ is maximum when $t=\mathrm{\infty };a(\mathrm{\infty })=0$

$a(t)$ is minimum when $t=0;a(0)=-x_0{\gamma }^2$

Note We should be careful about nature of variation of the curve and maximum and minimum value will be decided accordingly.

Thinking Process

In this problem we have to use the concept of relative velocity it will be subtracted for velocities in same direction and added for velocities in opposite directions.

Answer Given, Speed of first car $=18km/h$

Speed of second car $=27km/h$

$\therefore$ Relative speed of each car w.r.t. each other

$=18+27=45km/h$

Distance between the cars $=36km$

$\therefore$ Time of meeting the cars $(t)=\frac{\text{ Distance between the cars }}{\text{ Relative speed of cars }}=\frac{36}{45}=\frac{4}{5}h=0.8h$

Speed of the bird $(v_b)=36km/h$

$\therefore$ Distance covered by the bird $=v_b\times t=36\times 0.8=28.8km$

Thinking Process

When the man runs on the roof-top the velocity will be horizontal. Then, acceleration will be vertically down ward taken as $g$ and then equations of kinematies will be applied.

Answer Given, horizontal speed of the man $(u_x)=9m/s$

Horizontal distance between the two buildings $=10m$

Height difference between the two buildings $=9m$ and

$g=10m/s^2$

Let the man jumps from point $A$ and land on the roof of the next building at point $B$. Taking motion in vertical direction,

or

$\begin{array}{rl}& y=ut+\frac{1}{2}a{t}^2\\ & 9=0\times t+\frac{1}{2}\times 10\times t^2\\ & 9=5t^2\\ & t=\sqrt{\frac{9}{5}}=\frac{3}{\sqrt{5}}\end{array}$

$\therefore$ Horizontal distance travelled $=u_x\times t=9\times \frac{3}{\sqrt{5}}=\frac{27}{\sqrt{5}}m$

$\approx 12m$

Horizontal distance travelled by the man is greater than $10m$, therefore, he will land on the next building.

Thinking Process

In this problem as ball is dropped, initial velocity will be taken as zero. We will apply equations involving one dimensional motion.

Answer For the ball dropped from the building, $u_1=0,u_2=40m/s$

Velocity of the dropped ball after time $t$,

$\begin{array}{rl}& v_1=u_1+gt\\ & v_1=gt\end{array}$

(downward)

For the ball thrown up, $u_2=40m/s$

Velocity of the ball after time $t$

$\begin{array}{rl}{v}_2& =u_2-gt\\ & =(40-gt)\end{array}$

(upward)

$\therefore$ Relative velocity of one ball w.r.t. another ball

$\begin{array}{rl}& =v_1-v_2\\ & =gt-[-(40-gt)]=40m/s\end{array}$

Note When we are applying equations for rectilinear motion we should carefully put up the signs for the physical quantities.

Thinking Process

In this problems we will use the concept of slope. Suppose the slope is $m$ then equation will come out to bey $=mx+c$.

Answer Given, initial velocity $=v_0$

Let the distance travelled in time $t=x_0$.

For the graph

$\tan \theta =\frac{v_0}{x_0}=\frac{v_0-v}{x}$

where, $v$ is velocity and $x$ is displacement at any instant of time $t$.

$\begin{array}{rlr}\text{ From Eq. (i) }& v_0-v& =\frac{v_0}{x_0}x\\ \Rightarrow & v& =\frac{-v_0}{x_0}x+v_0\end{array}$

We know that

$\begin{array}{rl}\text{ Acceleration }a& =\frac{dv}{dt}=\frac{-v_0}{x_0}\frac{dx}{dt}+0\\ a& =\frac{-v_0}{x_0}(v)\\ & =\frac{-v_0}{x_0}(\frac{-v_0}{x_0}x+v_0)\\ & =\frac{v_0^2}{x_0^2}x-\frac{v_0^2}{x_0}\end{array}$

Graph of a versus $x$ is given above.

Thinking Process

In this problem equation of motion and Newton’s seconds law that is $F_{\text{ext }}=\frac{dp}{dt}$ will be used. where $dp$ is change in momentum over timedt.

Answer Given, height $(h)=1km=1000m$

$g=10m/s^2$

(a) Velocity attained by the rain drop in freely falling through a height $h$.

$\begin{array}{rl}V& =\sqrt{2gh}=\sqrt{2\times 10\times 1000}=100\sqrt{2}m/s\\ & =100\sqrt{2}\times \frac{60\times 60}{1000}km/h\\ & =360\sqrt{2}km/h\approx 510km/h\end{array}$

(b) Diameter of the drop $(d)=2r=4mm$

$\therefore$ Radius of the drop $(r)=2mm=2\times {10}^{-3}m$

Mass of a rain drop $(m)=V\times \rho$

$\begin{array}{rl}& =\frac{4}{3}\pi r^3\rho \\ & =\frac{4}{3}\times \frac{22}{7}\times (2\times {10}^{-3}{)}^3\times {10}^3(\because \text{ Density of water }={10}^{3}kg/m^3)\\ & \approx 3.4\times {10}^{-5}kg\end{array}$

Momentum of the rain $drop(p)=mv$

$\begin{array}{rl}& =3.4\times {10}^{-5}\times 100\sqrt{2}\\ & =4.7\times {10}^{-3}kg-m/s\\ & \approx 5\times {10}^{-3}kg-m/s\end{array}$

(c) Time required to flatten the drop = time taken by the drop to travel the distance equal to the diameter of the drop near the ground

$\begin{array}{rl}t& =\frac{d}{V}=\frac{4\times {10}^{-3}}{100\sqrt{2}}=0.028\times {10}^{-3}s\\ & =2.8\times {10}^{-5}s\approx 30ms\end{array}$

(d) Force exerted by a rain drop

$\begin{array}{rl}F& =\frac{\text{ Change in momentum }}{\text{ Time }}=\frac{p-0}{t}\\ & =\frac{4.7\times {10}^{-3}}{2.8\times {10}^{-5}}\approx 168N\end{array}$

(e) Radius of the umbrella $(R)=\frac{1}{2}m$

$\therefore$ Area of the umbrella $(A)=\pi R^2=\frac{22}{7}\times (\frac{1}{2}{)}^2=\frac{22}{28}=\frac{11}{14}\approx 0.8m^2$

Number of drops striking the umbrella simultaneously with average separation of $5cm$ $=5\times {10}^{-2}m$.

$=\frac{0.8}{(5\times {10}^{-2}{)}^2}=320$

$\therefore$ Net force exerted on umbrella $=320\times 168=53760N\approx 54000N$

Note In practice, the velocity of the drops decreases due to air friction.

Answer In this problem equations related to one dimensional motion will be applied for acceleration positive sign will be used and for retardation negative sign will be used.

Given, speed of car as well as truck $=72km/h$

Retarded motion for truck $v=u+a_{t}t$

$=72\times \frac{5}{18}m/s=20m/s$

or

$0=20+a_t\times 5$

$a_t=-4m/s^2$

Retarded motion for the car

$\begin{array}{rl}v& =u+a_{c}t\\ 0& =20+a_c\times 3\\ a_c& =-\frac{20}{3}m/s^2\end{array}$

$\text{ or }$

Let car be at a distance $x$ from truck, when truck gives the signal and $t$ be the time taken to cover this distance.

As human response time is $0.5s$, therefore, time of retarded motion of car is $(t-0.5)s$.

Velocity of car after time $t$,

$\begin{array}{rl}{v}_c& =u-at\\ & =20-(\frac{20}{3})(t-0.5)\end{array}$

Velocity of truck after time $t$

$v_t=20-4t$

To avoid the car bump onto the truck,

$\begin{array}{rl}{v}_c& =v_t\\ 20-\frac{20}{3}(t-0.5)& =20-4t\\ 4t& =\frac{20}{3}(t-0.5)\\ t& =\frac{5}{3}(t-0.5)\\ 3t& =5t-2.5\\ t& =\frac{2.5}{2}=\frac{5}{4}s\end{array}$

Distance travelled by the truck in time $t$,

$\begin{array}{rl}{s}_t& =u_{t}t+\frac{1}{2}{a}_t{t}^2\\ & =20\times \frac{5}{4}+\frac{1}{2}\times (-4)\times (\frac{5}{4}{)}^2\\ & =21.875m\end{array}$

Distance travelled by car in time $t=$ Distance travelled by car in $0.5s$ (without retardation)

• Distance travelled by car in $(t-0.5)s$ (with retardation)

$\begin{array}{rl}{s}_c& =(20\times 0.5)+20(\frac{5}{4}-0.5)-\frac{1}{2}(\frac{20}{3})(\frac{5}{4}-0.5{)}^2\\ & =23.125m\\ \therefore s_c-s_t& =23.125-21.875=1.250m\end{array}$

Therefore, to avoid the bump onto the truck, the car must maintain a distance from the truck more than $1.250m$.

Answer It this problem to calculate maximum velocity we will use $\frac{dv}{dt}=0$, then the time corresponding to maximum velocity will be obtained.

Given velocity

$v(t)=2t(3-t)=6t-2t^2$

(a) For maximum velocity $\frac{dv(t)}{dt}=0$

$\begin{array}{cccc}\Rightarrow & & \frac{d}{dt}(6t-2t^2)& =0\\ \Rightarrow & 6-4t& =0\\ \Rightarrow & t& =\frac{6}{4}=\frac{3}{2}s=1.5s\end{array}$

(b) From Eq. (i) $v=6t-2t^2$

$\begin{array}{cc}\Rightarrow & \frac{ds}{dt}=6t-2t^2\\ \Rightarrow & ds=(6t-2t^2)dt\end{array}$

where, $s$ is displacement

$\therefore$ Distance travelled in time interval 0 to $3s$.

$\begin{array}{rl}S& ={\int }_0^3(6t-2t^2)dt\\ & =[\frac{6t^2}{2}-\frac{2t^3}{3}{]}_0^3=[3t^2-\frac{2}{3}{t}^3{]}_0^3\\ & =3\times 9-\frac{2}{3}\times 3\times 3\times 3\\ & =27-18=9m.\\ \text{ Average velocity }& =\frac{\text{ Distance travelled }}{\text{ Time }}\\ & =\frac{9}{3}=3m/s\\ \Rightarrow x& =6t-2t^2\\ \Rightarrow 3& =6t-2t^2\\ \Rightarrow 2t^2-6t-3& =0\\ \Rightarrow t& =\frac{6\pm \sqrt{6^2-4\times 2\times 3}}{2\times 2}=\frac{6\pm \sqrt{36-24}}{4}\\ \Rightarrow & =\frac{6\pm \sqrt{12}}{4}=\frac{3\pm 2\sqrt{3}}{2}\end{array}$

Considering positive sign only

$t=\frac{3+2\sqrt{3}}{2}=\frac{3+2\times 1.732}{2}=\frac{9}{4}s$

(c) In a periodic motion when velocity is zero acceleration will be maximum putting $v=0$ in Eq. (i)

$0=6t-2t^2$

$\Rightarrow 0=t(6-2t)$

$\Rightarrow =t\times 2(3-t)=0$

t=0 or 3 s

(d) Distance covered in 0 to $3s=9m$

Distance covered in 3 to $6s={\int }_3^6(18-9t+t^2)dt$

$\begin{array}{rl}& =(18t-\frac{9t^2}{2}+\frac{t^3}{3}{)}_3^6\\ & =18\times 6-\frac{9}{2}\times 6^2+\frac{6^3}{3}-(18\times 3-\frac{9\times 3^2}{2}+\frac{3^3}{3})\\ & =108-9\times 18+\frac{6^3}{3}-18\times 3+\frac{9}{2}\times 9-\frac{27}{3}\\ & =108-18\times 9+\frac{216}{3}-54+4.5\times 9-9=-4.5m\end{array}$

$\therefore$ Total distance travelled in one cycle $=s_1+s_2=9-4.5=4.5m$

Number of cycles covered in total distance to be covered $=\frac{20}{4.5}\approx 4.44\approx 5$.

Thinking Process

In this question equation of kinematics will be applied with proper sign and to calculate time interval we will take difference of displacements.

Answer Let the speeds of the two balls ( 1 and 2 ) be $v_1$ and $v_2$ where

if $v_1=2v,v_2=v$

if $y_1$ and $y_2$ and the distance covered by

the balls 1 and 2 , respectively, before coming to rest, then

$y_1=\frac{v_1^2}{2g}=\frac{4v^2}{2g}\text{ and }{y}_2=\frac{v_2^2}{2g}=\frac{v^2}{2g}$

Since,

or

$y_1-y_2=15m,\frac{4v^2}{2g}-\frac{v^2}{2g}=15m\text{ or }\frac{3v^2}{2g}=15m$

or $v=10m/s$

Clearly, $v_1=20m/s$ and $v_2=10m/s$

$\begin{array}{rl}& y_1=\frac{v_1^2}{2g}=\frac{(20m{)}^2}{2\times 10m15}=20m\\ & y_2=y_1-15m=5m\end{array}$

If $t_2$ is the time taken by the ball 2 toner

a distance of $5m$, then from $y_2=v_{2}t-\frac{1}{2}g{t}_2^2$

$5=10t_2-5t_2^2\text{ or }{t}_2^2-2t_2+1=0$

where $t_2=15$

Since $t_1$ (time taken by ball 1 to cover distance of $20m$ ) is $2s$, time interval between the two thrvws

$=t_1-t_2=2s-1s=1s$

Note We should be very careful, when we are applying the equation of rectilinear motion.