Question

"a car moves with a speed of 40km/h can be stopped by applying brakes in 4m.if the same car is moving with a speed of 80km/h,what is the minimum stopping distance assuming that the retardation is constant"

Answer 1

Answer on Question ##69471 -Physics / Other

A car moves with a speed of $v_{1} = 40 \, \mathrm{km/h}$ can be stopped by applying brakes in $S_{1} = 4 \, \mathrm{m}$. If the same car is moving with a speed of $v_{2} = 80 \, \mathrm{km/h}$, what is the minimum stopping distance assuming that the retardation is constant.

Solution

$$S = \frac {v _ {f} ^ {2} - v _ {i} ^ {2}}{2 a}$$ $$v _ {f} = 0, \quad a < 0$$ $$S _ {1} = \frac {v _ {1 i} ^ {2}}{2 a};$$ $$S _ {2} = \frac {v _ {2 i} ^ {2}}{2 a}$$

Thus:

$$\frac {S _ {2}}{S _ {1}} = \frac {v _ {2 i} ^ {2}}{v _ {1 i} ^ {2}}$$ $$S _ {2} = \frac {v _ {2 i} ^ {2}}{v _ {1 i} ^ {2}} \times S _ {1} = \left(\frac {80}{40}\right) ^ {2} \times 4 = 16 \, \mathrm{m}.$$

Answers: $S_{2} = 16 \, \mathrm{m}$

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