A truck covers $40.0 m$ in $8 s$ while smoothly slowing down to a final speed of $2 m / s$ . Find its acceleration if it starts with a speed $10 m / s$ .

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Solution

Given:
Initial velocity $(u) = 10 m / s$
Distance covered by the car: $s = 40 m$
Final velocity of the car: $v = 2 m / s$
Time taken: $t = 8 s e c$

Acceleration $(a) = ?$
Using the equation of motion, $v = u + a t$
$a = \frac{v - u}{t} = \frac{2 - 10}{8} = - 1 m / s^{2}$

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A truck covers 40.0m in 8 s while smoothly slowing down to a final speed of 2m/s. Find its acceleration if it starts with a speed 10 m/s.

Given:Initial velocity (u)=10 m/sDistance covered by the car: s=40 mFinal velocity of the car: v=2 m/sTime taken: t=8 secAcceleration (a)=?Using the equation of motion, v=u+ata=v−ut=2−108=−1 m/s2

A truck covers $40.0 m$ in $8 s$ while smoothly slowing down to a final speed of $2 m / s$ . Find its acceleration if it starts with a speed $10 m / s$ .

Given:
Initial velocity $(u) = 10 m / s$
Distance covered by the car: $s = 40 m$
Final velocity of the car: $v = 2 m / s$
Time taken: $t = 8 s e c$

Acceleration $(a) = ?$
Using the equation of motion, $v = u + a t$
$a = \frac{v - u}{t} = \frac{2 - 10}{8} = - 1 m / s^{2}$