Answer to Question #94303 in Mechanics | Relativity for Maha

Question #94303
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 ?
1
Expert's answer
2019-09-12T10:37:08-0400

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}".


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