Question

A truck traveling with an initial velocity of 22m/s comes to a stop in 17.32 secs. What is the acceleration of the truck?

Answer 1

Explanation:

Given:

v₀ = 22 m/s

v = 0 m/s

t = 17.32 s

Find: a

v = at + v₀

(0 m/s) = a (17.32 s) + (22 m/s)

a = -1.270 m/s²

Round as needed.

Answer 2

Answer:

A truck traveling with an initial velocity of 22 m/s comes to a stop in 17.32 sec, the acceleration of the truck is $- 1.27 m/s^2$

Explanation

From the Given statements, we know that Initial Velocity (u) = 22 m/s

time (t) = 17.32 seconds

Final Velocity (v) = 0 m/s [ because the truck Stops]

find acceleration on Applying the Equation of Motion

v = u + at

substituting the values in above equation, we find

  a   =$\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}$

       =$\frac{0-22}{17.32} \mathrm{m} / \mathrm{s}^{2}$

 a  = $- 1.27 m/s^2$ [negative sign shows reverse direction]