Question

An object is moving with an initial velocity of 19 m/s.It is then subject to a constant acceleration of 2.5 m/s for 15s.How far will it have traveled during the time of its acceleration?

Answer 1

Answer:

566.3 m

Explanation:

The distance travelled by the object can be found by using the SUVAT equation:

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

where

u is the initial velocity

t is the time

a is the acceleration

For the object in this problem:

u = 19 m/s

a = 2.5 m/s^2

Substituting t = 15 s, we find the distance travelled:

$d=(19)(15)+\frac{1}{2}(2.5)(15)^2=566.3 m$