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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DMedium2 minQB

PhysicsMediumnumerical

An object loses $20%$ of its speed after running 108 m . If maximum distance travelled by the object before stopping is 100 x (in meter). Then value of x is.

Answer:

3.00

Solution:

$(\frac{4 V _{0}}{5})^{2} = V_{0}^{2} + 2 a \times 108$

$- \frac{9 V _{0}^{2}}{25} = 2 a (108)$

$2 a = \frac{- 9 V _{0}^{2}}{25 \times 108}$

For stopping $V = 0$

$O^{2} = V_{0}^{2} + 2 aS$

$S = \frac{- V _{0}^{2}}{2 a} = \frac{25 \times 108}{9} = 300$

$100 x = 300$ $x = 3$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: QB

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