(D) Given:Initial velocity $u = 200\; m/s$Final velocity $v = 100\; m/s$Distance $s = 10\; cm = 0.1\; m$Using the third equation of motion: $v^2 = u^2

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(D) Given:Initial velocity $u = 200\; m/s$Final velocity $v = 100\; m/s$Distance $s = 10\; cm = 0.1\; m$Using the third equation of motion: $v^2 = u^2 + 2as$$a = \frac{v^2 - u^2}{2s}$$a = \frac{(100)^2 - (200)^2}{2 \times 0.1}$$a = \frac{10000 - 40000}{0.2} = \frac{-30000}{0.2} = -150000\; m/s^2$Retardation is the magnitude of negative acceleration,so retardation $= 15 \times 10^4\; m/s^2$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

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The velocity of a bullet is reduced from $200\; m/s$ to $100\; m/s$ while travelling through a wooden block of thickness $10\; cm$. The retardation,assuming it to be uniform,will be ........... $\times 10^4\; m/s^2$.

A
$10$
B
$12$
C
$13.5$
D
$15$

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Motion in Straight Line

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Uniformly Accelerated Motion

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The acceleration of an object increases with time as $a = bt$. The object starts from the origin with an initial velocity $v_0$. Find the distance traveled by the object in time $t$.

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An automobile travelling with a speed of $60\,km/h$ can brake to stop within a distance of $20\,m$. If the car is going twice as fast,i.e.,$120\,km/h$,the stopping distance will be ........... $m$.

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$A$ car moving with a velocity of $20 \,m \,s^{-1}$ is stopped in a distance of $40 \,m$. If the same car is travelling at double the velocity, the distance travelled by it for the same retardation is (in $\,m$)

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The velocity of a bullet is reduced from $200\; m/s$ to $100\; m/s$ while travelling through a wooden block of thickness $10\; cm$. The retardation,assuming it to be uniform,will be ........... $\times 10^4\; m/s^2$.

Similar Questions

Consider that a truck is moving initially with $54 \,km/h$. It is stopped by the driver after looking at an obstacle with a deceleration of $10 \,m/s^2$. The distance travelled by the truck before coming to rest is (in $\,m$)

$A$ particle moves along a straight line in such a way that its acceleration is increasing at the rate of $2 m/s^3$. Its initial acceleration and velocity were $0$. The distance covered by it in $t = 3 s$ is ........ $m$.

The acceleration of an object increases with time as $a = bt$. The object starts from the origin with an initial velocity $v_0$. Find the distance traveled by the object in time $t$.

An automobile travelling with a speed of $60\,km/h$ can brake to stop within a distance of $20\,m$. If the car is going twice as fast,i.e.,$120\,km/h$,the stopping distance will be ........... $m$.

$A$ car moving with a velocity of $20 \,m \,s^{-1}$ is stopped in a distance of $40 \,m$. If the same car is travelling at double the velocity, the distance travelled by it for the same retardation is (in $\,m$)

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