Answer to Question #149053 in Physics for Melanie Belandres

Question

1. The makers of a certain automobile advertise that it will accelerate from 15 to 50 mi/hr in 13 s. Compute
a) the acceleration in m/s2
b) the distance the car travels in this time, assuming the acceleration to be constant.

Answer 1

Let the initial speed of the car be $v_i = 15mi/h\approx 6.7\space m/s$ and the final speed be $v_f = 50mi/h \approx 22.4m/s$. By definition, the acceleration is (here $t = 13s$ ):

$$a = \dfrac{v_f-v_i}{t} =\dfrac{22.4-6.7}{13} \approx 1.2m/s^2$$

The distance the car travels in this time, assuming the acceleration to be constant:

$$d = v_it + \dfrac{at^2}{2} = 6.7\cdot 13 + \dfrac{1.2\cdot 13^2}{2} =188.5m$$

Answer. $a = 1.2m/s^2, \space d = 188.5 m.$