Answer to Question #176537 in Physics for Chimka

Question

The speed if a car increased from 36km/h to 96km/h in 0.6km.how long did this take and what was the acceleration

Answer 1

The acceleration of the car by definition is:

$a = \dfrac{v_f-v_i}{t}$

where $v_f = 96km/h = \dfrac{80}{3}m/s$ is the final speed of the car, $v_i = 36km/h = 10m/s$ is the initial speed of the car, and $t$ is the time of acceleration.

The distance $d = 0.6km = 600m$ travelled under the constant acceleration is given by the kinematic equation:

$d = v_it+\dfrac{at^2}{2}$

Substituting $a$, obtain:

$d = v_it + \dfrac{v_ft}{2} - \dfrac{v_it}{2}= \dfrac{(v_f+v_i)t}{2}\\ t = \dfrac{2d}{v_f+v_i} \\ t = \dfrac{2\cdot 600}{80/3+10}\approx 32.7s$

Answer. 32.7 s