The brakes applied to a car produce an acceleration of 10 m/s2 in the opposite direction to the motion.if it takes 5 s to stop after the application of brakes, then how much distance is covered by the car before coming to rest?

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Solution

Here final velocity v=0, hence the total distance travelled will be $S = - \frac{1}{2} a t^{2}$ . Here a=-10 m/s² and time t=5 s. So the distance travelled is $S = - \frac{1}{2} \times (- 10) \times 5^{2} = 125$ m.

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The brakes applied to a car produce an acceleration of 10 m/s2 in the opposite direction to the motion.if it takes 5 s to stop after the application of brakes, then how much distance is covered by the car before coming to rest?

Here final velocity v=0, hence the total distance travelled will be S=-12at2. Here a=-10 m/s2 and time t=5 s. So the distance travelled is S=-12×-10×52=125 m.

Here final velocity v=0, hence the total distance travelled will be $S = - \frac{1}{2} a t^{2}$ . Here a=-10 m/s² and time t=5 s. So the distance travelled is $S = - \frac{1}{2} \times (- 10) \times 5^{2} = 125$ m.