Answer to Question #94303 in Mechanics | Relativity for Maha

Question

A motorcycle, initially moving with a speed of 4.0 m/s , speeds up at a constant rate to 5.0 m/s after covering a distance of 5.0 m . What is the acceleration of the motorcycle ?

Answer 1

Let us write the equation of motion of the motorcycle uniformly accelerating from $v_0$ to $v_1$and covering the distance S: $v_1 = v_0 + a t$ , $S = v_0 t + \frac{a t^2}{2}$ . From the first equation, the time needed for that motion is $t = \frac{v_1 - v_0}{a}$ . Let us substitute the last expression into the expression for $S$: $S = \frac{v_0 (v_1 - v_0)}{a} + \frac{(v_1-v_0)^2}{2 a^2} = \frac{1}{2 a}(v_1^2 - v_0^2)$ . Hence, from the last formula, the acceleration is $a = \frac{v_1^2 - v_0^2}{2 S} = 0.9 \frac{m}{s}$.