Question

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?
B) What total distance would you travel from when you first apply the brakes until the car stops?

Answer 1

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity$=70 mph\approx 31.29 m/s$

velocity after 5 s is $30 mph\approx 13.41 m/s$

Therefore acceleration during these 5 s

$a=\frac{v-u}{t}$

$a=\frac{13.41-31.29}{5}=-3.576 m/s^2$

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

$0=31.29-3.576\times t$

$t=\frac{31.29}{3.576}=8.75 s$

(b)total distance traveled before stoppage

$v^2-u^2=2as$

$0^2-31.29^2=2\times (-3.576)\cdot s$

s=136.89 m

Answer 2

A. The time taken for the car to stop is 8.75 s

B. The distance travelled when the brakes were applied till the car stops is 136.89 m

A. Determination of the time taken for the car to stop.

• We'll begin by calculating the deceleration of the car

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

Deceleration (a) =?

$a \: = \frac{v \: - u}{t} \\ \\ a = \frac{13.41 - 31.29}{5} \\ \\ a \: = \frac{ - 17.88}{5}$

a = –3.576 m/s²

• Finally, we shall determine the time taken for the car to stop.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Time (t) =?

$v \: = u \: + at \\ 0 \: = 31.29 \: + \: ( - 3.576 \times t) \\ 0 \: = 31.29 \: - 3.576 \times t \\ collet \: like \: terms \\ 0 - 31.29 \: = - 3.576 \times t \\ - 31.29 \: = - 3.576 \times t \\ divide \: both \: side \: by \: - 3.576 \\ t \: = \frac{- 31.29}{- 3.576}$

t = 8.75 s

Thus, the time taken for the car to stop is 8.75 s

B. Determination of the total distance travelled when the brakes were applied.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Distance (s) =?

${v}^{2} = {u}^{2} + 2as \\ {0}^{2} = {31.29}^{2} + (2 \times - 3.576 \times s) \\ 0 = 979.0641 - 7.152 s \\ collect \: like \: terms \\ 0 - 979.0641 = - 7.152 s \\ - 979.0641 = - 7.152 s \\ divide \: both \: side \: by \: - 7.152 \\ s = \frac{- 979.0641}{- 7.152}$

s = 136.89 m

Therefore, the total distance travelled by the car when the brakes were applied is 136.89 m

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