Question

1. Ahmad is driving North on Highway 69 at 90 km/h and sees a large moose on the road. He quickly slams on his brakes, but his reaction time is 0.85 s (as he sees the moose, thinks about his response, and then presses the brake pedal). He presses the brake for 3.5 s and comes to a stop just in time.
a) Find the distance travelled after seeing the moose and before pressing the brake.
b) Find the total distance he travelled before coming to a stop.
c) Find the average acceleration once he presses the brake.

Answer 1

Answer on Question # 71584, Physics / Mechanics | Relativity

Question

1. Ahmad is driving North on Highway 69 at $90~\mathrm{km/h}$ and sees a large moose on the road. He quickly slams on his brakes, but his reaction time is $0.85~\mathrm{s}$ (as he sees the moose, thinks about his response, and then presses the brake pedal). He presses the brake for $3.5~\mathrm{s}$ and comes to a stop just in time.

a) Find the distance travelled after seeing the moose and before pressing the brake.

b) Find the total distance he travelled before coming to a stop.

c) Find the average acceleration once he presses the brake.

Solution

a) Find the distance travelled after seeing the moose and before pressing the brake, $x_{1}$. Use the formula for distance at constant velocity

$$x_{1} = v_{0} t$$

Plug $v_{0} = 90~\mathrm{km/h}$ and $t = 0.85~\mathrm{s}$

$$x_{1} = v_{0} t = 90~\mathrm{km/h} \times 0.85~\mathrm{s}$$

Since the units are mixed, we need to convert everything into SI units of meters and seconds.

$$x_{1} = 90~\frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000~\mathrm{m}}{1~\mathrm{km}} \times \frac{1~\mathrm{h}}{3600~\mathrm{s}} \times 0.85~\mathrm{s} = 25~\mathrm{m/s} \times 0.85~\mathrm{s} = 21.25~\mathrm{m}$$

b) Find the distance from the beginning of the braking to the stop, $x_{2}$. Use the formula for distance at constant acceleration

$$x_{2} = \bar{v} t$$

where

$$\bar{v} = \frac{v_{0} + v}{2}$$

is the average velocity, that is average of the initial and final velocities. Since $v_{0} = 90~\mathrm{km/h} = 25~\mathrm{m/s}$ and $v = 0~\mathrm{m/s}$ we get

$$\bar{v} = \frac{25 + 0}{2} = 12.5~\mathrm{m/s}$$

Now we find $x_{2}$ plugging $\bar{v} = 12.5~\mathrm{m/s}$ and $t = 3.5~\mathrm{s}$

$$x_{2} = \bar{v} t = 12.5~\mathrm{m/s} \times 3.5~\mathrm{s} = 43.75~\mathrm{m}$$

Find the total distance he travelled before coming to a stop

$$x = x_{1} + x_{2} = 21.25~\mathrm{m} + 43.75~\mathrm{m} = 65~\mathrm{m}$$

c) Find the average acceleration once he presses the brake. Use the formula for average acceleration

$$\bar{a} = \frac{v_{f} - v_{0}}{\Delta t}$$

where $v_{f}$ is the final velocity, $v_{f} = 0$, $v_{0}$ is the initial velocity, $\Delta t$ is a time from the beginning of the braking to the stop, $\Delta t = 3.5~\mathrm{s}$. Plugging these numbers, we obtain

$$\bar {a} = \frac {0 - 25}{3.5} = - \frac {50}{7} = - 7.14 \ \mathrm{m/s^2}$$

Answer.

a) The distance travelled after seeing the moose and before pressing the brake is 21.25 m.

b) The total distance he travelled before coming to a stop is 65 m.

c) The average acceleration once he presses the brake is $\bar{a} = -7.14\ \mathrm{m/s^2}$.

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