Question

A motorcycle has a constant acceleration of 2.5 m/s2. How much time is required for the motorcycle to change its speed from
a) 20 m/s to 30 m/s?
b) 50 m/s to 60 m/s?
c) How far does it travel in each case?

Answer 1

a) We can find the time required for the motorcycle to change its speed from 20 m/s to 30 m/s from the kinematic equation:

$$v=v_0+at_1,$$ $$t_1=\dfrac{v-v_0}{a}=\dfrac{30\ \dfrac{m}{s}-20\ \dfrac{m}{s}}{2.5\ \dfrac{m}{s^2}}=4.0\ s.$$

b) We can find the time required for the motorcycle to change its speed from 50 m/s to 60 m/s from the same kinematic equation:

$$v=v_0+at_2,$$ $$t_2=\dfrac{v-v_0}{a}=\dfrac{60\ \dfrac{m}{s}-50\ \dfrac{m}{s}}{2.5\ \dfrac{m}{s^2}}=4.0\ s.$$

c) We can find the distance traveled by the motorcycle in each cases from the kinematic equation:

$$d_1=v_0t_1+\dfrac{1}{2}at_1^2,$$ $$d_1=20\ \dfrac{m}{s}\cdot 4.0\ s+\dfrac{1}{2}\cdot 2.5\ \dfrac{m}{s^2}\cdot (4.0\ s)^2=100\ m,$$ $$d_2=v_0t_2+\dfrac{1}{2}at_2^2,$$ $$d_2=50\ \dfrac{m}{s}\cdot 4.0\ s+\dfrac{1}{2}\cdot 2.5\ \dfrac{m}{s^2}\cdot (4.0\ s)^2=220\ m.$$

Answer:

a) $t_1=4.0\ s.$

b) $t_2=4.0\ s.$

c) $d_1=100\ m, d_2=220\ m.$