Question

A car is traveling at 40 mi/h when the driver slams on the brakes coming to rest over a distance of 68 meters. What is the acceleration of the car in m/s2?

Answer 1

Answer on Question #66359, Physics / Optics

Question:

A car is traveling at $40\mathrm{mi/h}$ when the driver slams on the brakes coming to rest over a distance of 68 meters. What is the acceleration of the car in $\mathrm{m/s^2}$?

Solution:

Let $v_{1}$ be the initial truck's speed, $v_{2}$ — its final speed, $S$ — distance, $a$ — acceleration of the truck, and $t$ — time spent for decreasing the speed.

For uniformly accelerated motion

$$v _ {2} = v _ {1} + a t$$ $$S = v _ {1} t + \frac {a t ^ {2}}{2}$$

From $1^{\mathrm{st}}$ equation $t = \frac{v_2 - v_1}{a}$

and then $S = v_{1}\frac{v_{2} - v_{1}}{a} +\frac{a\left(\frac{v_{2} - v_{1}}{a}\right)^{2}}{2} = v_{1}\frac{v_{2} - v_{1}}{a} +\frac{(v_{2} - v_{1})^{2}}{2a} = \frac{v_{2}^{2} - v_{1}^{2}}{2a}.$

Finally $a = \frac{v_2^2 - v_1^2}{2S}$

$$v _ {1} = 4 0 \frac {m i}{h} = 4 0 \cdot \frac {1 6 0 9 . 3 4 m}{3 6 0 0 s} = 1 7. 8 8 \frac {m}{s}, v _ {2} = 0 \frac {m}{s}, S = 6 8 m$$ $$a = \frac {0 ^ {2} - 1 7 . 8 8 ^ {2}}{2 \cdot 6 8} = - 2. 3 5 \frac {m}{s ^ {2}}$$

Answer:

$$- 2. 3 5 \frac {m}{s ^ {2}}$$

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