Question

A motorcycle’s brakes are designed to absorb 35 kJ/s. The motorcycle and rider have a mass of 280 kg and are travelling at 100 km/h. How fast will they be going after 3 seconds braking

Answer 1

Answer on Question#45015, Physics, Mechanics | Kinematics | Dynamics

Let us denote given coefficient of absorption as $k$ . Since $\frac{\Delta E}{\Delta t} = -k$ , and difference of energies is the difference of kinetic energies $\Delta E = \frac{mv_2^2}{2} - \frac{mv_1^2}{2}$ , let us find $v_2: v_2 = \sqrt{\frac{-2}{m}k\Delta t + v_1^2}$ . Substituting $k = 35 \cdot 10^3\frac{J}{s}, m = 280kg, v_1 = 100\frac{km}{h} = \frac{250}{9}\frac{m}{s}, \Delta t = 3s$ , obtain $v_2 \approx 4.65\frac{m}{s}$ or $v_2 \approx 16.74\frac{km}{h}$ .

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