Question

Assume you are driving 20 mph on a straight road. Also, assume that at a speed of 20 miles per hour, it takes 100 feet to stop. If you were to increase your speed to 60 miles per hour, your stopping distance is now:

Answer 1

Answer:900  feet

Explanation:

Given

Velocity $\left ( V_1\right )=20 mph\approx 29.334 ft/s$

it take 100 feet to stop

Using Equation of motion

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

where

v,u=Final and initial velocity

a=acceleration

s=distance moved

$0-\left ( 29.334\right )^2=2\left (-a\right )\left ( 100\right )$

$a=\frac{29.334^2}{2\times 100}=4.302 ft/s^2$

When velocity is 60 mph$\approx 88.002 ft/s$

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

$0-\left ( 88.002\right )^2=2\left ( -4.302\right )\left ( s\right )$

s=900.08 feet

Answer 2

Answer:

900 feet          

Explanation:

Initial Speed, u₁ = 20 mph

Stopping distance, s₁ = 100 feet

Initial Speed, u₂ = 60 mph

Then, the stopping distance can be calculated using the third equation of motion:

$s=\frac{v^2-u^2}{2a}$

There would be same acceleration and final velocity would be zero (v=0).

$s=\frac{0-u^2}{2a}\\ \frac{s_2}{s_1}=\frac{u_2^2}{u_1^2}\\s_2= 100 ft\frac{(60)^2}{(20)^2} =900 feet$