Question

Upon entering a straight section of highway, Batman accelerates the Batcycle for 7.50 s, increasing its velocity from 24 m/s [S] to 38 m/s [S]. Calculate the Batcycle’s displacement while accelerating. 

Answer 1

Let's first find the acceleration of the Batcycle:

$$a=\dfrac{v-v_0}{t}=\dfrac{38\ \dfrac{m}{s}-24\ \dfrac{m}{s}}{7.5\ s}=1.86\ \dfrac{m}{s^2}.$$

Finally, we can find the Batcycle’s displacement from the kinematic equation:

$$d=v_0t+\dfrac{1}{2}at^2,$$ $$d=24\ \dfrac{m}{s}\cdot 7.5\ s+\dfrac{1}{2}\cdot 1.86\ \dfrac{m}{s^2}\cdot(7.5\ s)^2=232.3\ m.$$

Answer:

$d=232.3\ m.$