SATHEE: Kinematics 4 Question 9

Kinematics 4 Question 9

9. A particle of mass $m$ moves on the $x$ -axis as follows : it starts from rest at $t = 0$ from the point $x = 0$ and comes to rest at $t = 1$ at the point $x = 1$ . No other information is available about its motion at intermediate times $(0 < t < 1)$ . If $α$ denotes the instantaneous acceleration of the particle, then

(1993, 2M)

(a) $α$ cannot remain positive for all $t$ in the interval $0 \leq t \leq 1$

(b) $| α |$ cannot exceed 2 at any point in its path

(d) $α$ must change sign during the motion, but no other assertion can be made with the information given

(1984, 2M)

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Answer:

Correct Answer: 9. (c)

Solution:

Since, the body is at rest at $x = 0$ and $x = 1$ . Hence, $α$ cannot be positive for all time in the interval $0 \leq t \leq 1$ .

Therefore, first the particle is accelerated and then retarded.

Now, total time $t = 1 s$ (given)

Total displacement, $s = 1 m$ (given)

$s = Area under v - t graph$

$∴$ Height or $v_{max} = \frac{2 s}{t} = 2 m / s$ is also fixed.

$[Area or s = \frac{1}{2} \times t \times v_{max}]$

If height and base are fixed, area is also fixed .

In case $2 :$ Acceleration $=$ Retardation $= 4 m / s^{2}$

In case 1 : Acceleration $> 4 m / s^{2}$ while

Retardation $< 4 m / s^{2}$

While in case 3 : Acceleration $< 4 m / s^{2}$ and

Retardation $> 4 m / s^{2}$

Hence, $| α | \geq 4$ at some point or points in its path.

Kinematics 4 Question 9

Answer:

Solution:

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Kinematics 4 Question 9

9. A particle of mass m moves on the x-axis as follows : it starts from rest at t=0 from the point x=0 and comes to rest at t=1 at the point x=1. No other information is available about its motion at intermediate times (0<t<1). If α denotes the instantaneous acceleration of the particle, then

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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