Question

An object moving with constant acceleration changes its speed from 20 m/s to 60 m/s in 2.0 s. What is the acceleration? (b) How far did it move in this time?

Answer 1

Answer:

acceleration, a = $20 m/s^{2}$

distance, d = 80 m

Given:

Initial velocity of object, v = 20 m/s

Final velocity of object, v' = 60 m/s

Time interval, $\Delta t = 2.0 s$

Solution:

(a) Acceleration is the rate at which velocity changes and constant acceleration is when the velocity changes by equal amount in equal intervals of time.

Thus

acceleration, a = $\frac{v' - v}{\Delta t}$

a = $\frac{60 - 20}{2.0} = 20 m/s^{2}$

(b) Now, the distance covered is  given by:

$d = vt + \frac{1}{2}at^{2}$

$d = 20\times 2.0 + \frac{1}{2}20\times 2^{2} = 80 m$