Question

A sports car accelerates in third gear from 48.5 km/h to 80.2 km/h in 3.6 s. (a) What is the average acceleration of the car? (in m/s^2)
(b) If the car maintained this acceleration after reaching 80.2 km/h, how fast would it be moving 4.0 seconds later? (in km/h)

Answer 1

Explanation:

It is given that,

Initial velocity of the car, u = 48.5 km/h = 13.47 m/s

Final velocity of the car, v = 80.2 km/h = 22.27 m/s

Time, t = 3.6 s

(a) Average acceleration is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{22.27\ m/s-13.47\ m/s}{3.6\ s}$

$a=2.45\ m/s^2$

(b) We need to find the final velocity after 4 seconds. From first equation of motion as :

v = u + a t

$v=22.27\ m/s+2.45\ m/s^2\times 4\ s$

v = 32.07 m/s

or

v = 115.45 km/h

Hence, this is the required solution.