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Standard XII

Physics

2nd Equation of Motion

At time t = 0...

Question

At time t = 0, a car moving along a straight line has a velocity of $16 m s^{- 1}$ . It slows down with an acceleration of -0.5 t $m s^{- 1}$ , where t is in second. Mark the correct statement(s).

The direction of velocity changes at t = 8s

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The distance travelled in 4 is approximately 58.67 m

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The distance travelled by the particle in 10 s is 94 m.

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The speed of particle at t = 10 s is 9

m s^{- 1}

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Solution

The correct option is A The direction of velocity changes at t = 8s
from given a = -0.5t
$\frac{d V}{d t} = - 0.5 t ⟹ V - U = \frac{- 0.5 t^{2}}{2}$
here the direction of velocity changes when V = 0
therefore V = 0 & U= $16 \frac{m}{s}$
now $0 - 16 = \frac{- 0.5 t^{2}}{2} ⟹ t^{2} = 64 ⟹ t = 8 s$

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At time t = 0, a car moving along a straight line has a velocity of 16 ms−1. It slows down with an acceleration of -0.5 t ms−1, where t is in second. Mark the correct statement(s).

The correct option is A The direction of velocity changes at t = 8sfrom given a = -0.5t dVdt=−0.5t⟹V−U=−0.5t22 here the direction of velocity changes when V = 0 therefore V = 0 & U=16msnow 0−16=−0.5t22⟹t2=64⟹t=8s

At time t = 0, a car moving along a straight line has a velocity of $16 m s^{- 1}$ . It slows down with an acceleration of -0.5 t $m s^{- 1}$ , where t is in second. Mark the correct statement(s).