Question

A bicycle starts at 2.5m/s and accelerates along a straight path to a speed of 12.5m/s in a time of 4.5 seconds. What is the bicyclist’s acceleration to the nearest tenth of a m/s^2 ?

Answer 1

Answer:

The bicyclist's acceleration is 2.2m/s^2

Explanation:

Given

$u = 2.5m/s$ ---- Initial Velocity

$v = 12.5m/s$ ---- Final Velocity

$t = 4.5s$ ---- Time

Required

Determine the acceleration

This will be solved using the first equation of motion

$v = u + at$

Substitute values for v, u and a

$12.5 = 2.5 + a * 4.5$

$12.5 = 2.5 + 4.5a$

Collect Like Terms

$4.5a = 12.5 - 2.5$

$4.5a = 10.0$

Solve for a

$a = 10.0/4.5$

$a = 2.2m/s^2$ ---- (approximated)

Hence, the bicyclist's acceleration is 2.2m/s^2