Respuesta :
Answer:900 feet
Explanation:
Given
Velocity [tex]\left ( V_1\right )=20 mph\approx 29.334 ft/s[/tex]
it take 100 feet to stop
Using Equation of motion
[tex]v^2-u^2=2as [/tex]
where
v,u=Final and initial velocity
a=acceleration
s=distance moved
[tex]0-\left ( 29.334\right )^2=2\left (-a\right )\left ( 100\right )[/tex]
[tex]a=\frac{29.334^2}{2\times 100}=4.302 ft/s^2[/tex]
When velocity is 60 mph[tex]\approx 88.002 ft/s[/tex]
[tex]v^2-u^2=2as [/tex]
[tex]0-\left ( 88.002\right )^2=2\left ( -4.302\right )\left ( s\right )[/tex]
s=900.08 feet
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:
[tex]s=\frac{v^2-u^2}{2a}[/tex]
There would be same acceleration and final velocity would be zero (v=0).
[tex]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 [/tex]
