Question

A car is traveling east at 15.0 m/s when it turns due north and accelerates to 40.0 m/s, all during a time of 9.00 s. Calculate the magnitude of the car's average acceleration (in m/s2).

Answer 1

Answer:

4.75 m/s^2

Explanation:

The rate of change of velocity is called acceleration.

initial velocity, u = 15 m/s along east

Final velocity, v = 40 m /s along north

time, t = 9 s

$\overrightarrow{u} = 15\widehat{i}$

$\overrightarrow{v} = 40\widehat{j}$

Change in velocity,

$\overrightarrow{v}-\overrightarrow{u}=40\widehat{j}-15\widehat{i}$

Acceleration,

$\overrightarrow{a}=\frac{40\widehat{j}-15\widehat{i}}{9}$

Magnitude of acceleration

$\frac{\sqrt{40^{2}+15^{2}}}{9}$ = 4.75 m/s^2