Question

a cart rolls down a track. it starts from rest and attains a speed of 2.7 m/s in 3 s, what is the acceleration and distance traveled?

Answer 1

Answer:

Acceleration and the distance traveled are 0.9 m/s² and 4.05 m

Explanation:

Given that,

Initail speed, u = 0

Final speed, v = 2.7 m/s

Time, t = 3 s

To find,

The acceleration and distance traveled.

Solution,

Acceleration is change in speed per unit time

$a=\dfrac{v-u}{t}\\\\a=\dfrac{2.7-0}{3}\\\\a=0.9\ m/s^2$

Let s is the distance,

$d=ut+\dfrac{1}{2}at^2\\\\\because u=0\\\\d=0+\dfrac{1}{2}\times 0.9\times 3^2\\\\d=4.05\ m$

So, acceleration and the distance traveled are 0.9 m/s² and 4.05 m