Answer

Diameter of circular track (D) $=200m$

Radius of circular track $(r)=200/2=100m$

Time taken by the athlete for one round $(t)=40s$

Distance covered by athlete in one round $(s)=2\pi r$

$=2\times (22/7)\times 100$

Speed of the athlete $(v)=$ Distance $/$ Time

$=(2\times 2200)/(7\times 40)$

$=4400/7\times 40$

Therefore, Distance covered in $140s=$ Speed $(s)\times$ Time $(t)$

$=4400/(7\times 40)\times (2\times 60+20)$

$=4400/(7\times 40)\times 140$ $=4400\times 140/7\times 40$

$=2200m$

Number of round in $40s=1$ round

Number of round in $140s=140/40$

$=3^{1/2}$

After taking start from position $X$, the athlete will be at postion $Y$ after $3\frac{1}{2}$ rounds as shown in figure

Hence, Displacement of the athlete with respect to initial position at $x=$ xy $=$ Diameter of circular track

$=200m$

Answer

Total Distance covered from $AB=300m$

Total time taken $=2\times 60+30s$

$=150s$

Therefore, Average Speed from AB = Total Distance / Total Time $=300/150m{s}^{-1}$

$=2m{s}^{-1}$

Therefore, Velocity from AB =Displacement AB $/$ Time $=300/150m{s}^{-1}$

$=2m{s}^{-1}$

Total Distance covered from $AC=AB+BC$

$=300+200m$

Total time taken from A to C = Time taken for AB + Time taken for BC $=(2\times 60+30)+60s$

$=210s$

Therefore, Average Speed from AC = Total Distance /Total Time

$=400/210m{s}^{-1}$

$=1.904m{s}^{-1}$

Displacement (S) from $A$ to $C=AB-BC$

$=300-100m$

$=200m$

Time (t) taken for displacement from AC $=210s$

Therefore, Velocity from AC = Displacement (s) / Time (t)

$=200/210m{s}^{-1}$

$=0.952m{s}^{-1}$

Answer

The distance Abdul commutes while driving from Home to School $=S$ Let us assume time taken by Abdul to commutes this distance $=t_1$ Distance Abdul commutes while driving from School to Home $=S$ Let us assume time taken by Abdul to commutes this distance $=t_2$ Average speed from home to school $v_{1av}=20km{h}_{\text{- }}$ Average speed from school to home $v_{2av}=30km{h}^{-1}$ Also we know Time taken form Home to School $t_1=S/v_{1av}$ Similarly Time taken form School to Home $t_2=S/V_{2av}$ Total distance from home to school and backward $=2S$ Total time taken from home to school and backward $(T)=S/20+S/30$ Therefore, Average speed $(V_{av})$ for covering total distance $(2S)=$ Total Dostance/Total Time

$=2S/(S/20+S/30)$

$=2S/[(30S+20S)/600]$

$=1200S/50S$

$=24km{h}^{-1}$

Answer

Given Initial velocity of motorboat, $u=0$

Acceleration of motorboat, $a=3.0m{s}^{-2}$

Time under consideration, $t=8.0s$

We know that Distance, $s=ut+(1/2)a{t}^2$

Therefore, The distance travel by motorboat $=0\times 8+(1/2)3.0\times 82$

$=(1/2)\times 3\times 8\times 8m$

$=96m$

Answer

(a) Object B

(b) No

(c) $5.71km$

(d) $5.14km$

(a) Speed $=\frac{\text{ Distance }}{\text{ Time }}$

Slope of graph $=\frac{y\text{-axis }}{x-\text{ axis }}=\frac{\text{ Distance }}{\text{ Time }}$

Therefore, Speed = slope of the graph

Since slope of object B is greater than objects A and C, it is travelling the fastest.

(b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.

(c)

Since there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals $4/7km$.

Since the initial point of an object C is 4 graph units away from the origin, Its initial distance from the origin is $4\times (4/7)km=16/7km$

When B passes A, the distance between the origin and C is $8km$

Therefore, total distance travelled by C in this time = $8–(16/7)km=5.71km$

(d) The distance that object B has covered at the point where it passes C is equal to 9 graph units. Therefore, total distance travelled by B when it crosses C = $9\times (4/7)=5.14km$

Answer

Let us assume, the final velocity with which ball will strike the ground be ’ $v$ ’ and time it takes to strike the ground be ’ $t$ ’ Initial Velocity of ball, $u=0$

Distance or height of fall, $s=20m$

Downward acceleration, $a=10m{s}^{-2}$

As we know, $2as=v^2-u^2$

$v^2=2as+u^2$

$=2\times 10\times 20+0$

$=400$

$\therefore$ Final velocity of ball, $v=20m{s}^{-1}$

$t=(v-u)/a$

$\therefore$ Time taken by the ball to strike $=(20-0)/10$

$=20/10$

$=2$ seconds

Answer

(a)

The shaded area which is equal to $1/2\times 4\times 6=12m$ represents the distance travelled by the car in the first $4s$.

(b)

The part of the graph in red colour between time $6s$ to $10s$ represents uniform motion of the car.

Answer

(a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) It is possible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time

Circular motion is an example of an object moving with acceleration but with uniform speed.

An object moving in a circular path with uniform speed is still under acceleration because the velocity changes due to continuous changes in the direction of motion.

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

Answer

Radius of the circular orbit, $r=42250km$

Time taken to revolve around the earth, $t=24h$

Speed of a circular moving object, $v=(2\pi r)/t$

$=[2\times (22/7)\times 42250\times 1000]/(24\times 60\times 60)$

$=(2\times 22\times 42250\times 1000)/(7\times 24\times 60\times 60)m{s}^{-1}$

$=3073.74m{s}^{-1}$