The brakes applied to a car produce an acceleration of $6\, m s ^{-2}$ in the opposite direction of motion. If the car takes $2\, s$ to stop after the application of the brakes, calculate the distance it travels during this time.

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Given $a=-6 m s ^{-2}, t=2 s , v=0 m s ^{-1}$

$(i)$ Using the equation $v=u+a t,$ we have

$u=v-a t=0-(-6) \times 2$

$=12 m s ^{-1}$

$(ii)$ Now, using the expression $S=u t+1 / 2 a t^{2}$

we have $S=12 \times 2+\frac{1}{2} \times-6 \times(2)^{2}=12 m$

Similar Questions

Draw velocity-time graphs for the following situations

$(i)$ When body is moving with uniform velocity.

$(ii)$ When body is moving with variable velocity, but uniform acceleration.

$(iii)$ When body is moving with variable velocity, but uniform retardation.

$(iv)$ When body is moving with a variable velocity and variable acceleration.

Under what condition will the displacement and distance have the same magnitude ?

What is the nature of the displacement$-$time graph of a body moving with constant velocity ?

A train starting from rest picks up a speed of $10\, m s ^{-1}$ in $100\, s$. It continues to move at the same speed for the next $250\, s$. It is then brought to rest in the nert $50\, s$. Plot a speed$-$time graph for the entire motion of the train.

$(i)$ acceleration of the train while accelerating,

$(ii)$ retardation of the train while retarding,

$(iii)$ and the total distance covered by the train.

Differentiate between distance and displacement.