Question

a 1600kg car driving at 80kmhr^-1 puts the brakes on. if µ =0.68, what's the stopping distance? (deceleration due to brakes =2.9ms^-2)

Answer 1

Answer on Question #72915, Physics / Mechanics | Relativity

Question. A $1600 \, kg$ car driving at $80 \, km/hr$ puts the brakes on. If $\mu = 0.68$, what's the stopping distance? (deceleration due to brakes $= 2.9 \, m/s^2$)?

Given. $m = 1600 \, kg$; $v_0 = 80 \, km/hr \approx 22.22 \, m/s$; $\mu = 0.68$; $a = -2.9 \, m/s^2$; $v_f = 0 \, m/s$.

Find. $s - ?$

Solution.

If a driver puts on the brakes of a car, the car will not come to a stop immediately. The stopping distance is the distance the car travels before it comes to a rest. It depends on the speed of the car and the coefficient of friction ($\mu$) between the wheels and the road. This stopping distance formula does not include the effect of anti-lock brakes or brake pumping. The $SI$ unit for stopping distance is meters.

$$s = \frac{v_f^2 - v_0^2}{2a} = \frac{0 - v_0^2}{2(-a)} = \frac{v_0^2}{2a}.$$

According to the Second Newton's Low

$$F_{fr} = ma \rightarrow \mu mg = ma \rightarrow a = \mu g$$

Finally

$$s = \frac{v_0^2}{2\mu g} = \frac{22.22^2}{2 \cdot 0.68 \cdot 9.8} = 37.044 \, m.$$

Answer. $s = 37.044 \, m$.

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