Question

A truck covers 40.0 m in 9.50 s while uniformly slowing down to a final velocity of 2.75 m/s.

a. Find its original speed.
b. Find its acceleration.

Answer 1

Step-by-step explanation:

Given that,

Distance covered, d = 40 m

Time, t = 9.5 s

Final velocity, v = 2.75 m/s

(a) Let u be the original speed of the truck. We can find it using first equation of motion.

`v=u+at\\\\2.75=u+2.75* 9.5\\\\2.75-26.125=u\\\\u=-23.375\ m/s`

(b) Acceleration = rate of change of velocity

`a=(v-u)/(t)\\\\a=(2.75-(-23.375))/(9.5)\\\\=2.75\ m/s^2`

So, the original speed is -23.375 and acceleration is 2.75 m/s².