Answer:
The distance the have traveled before come into the rest = 72.72 ft
Step-by-step explanation:
Velocity of the car u = 40 [tex]\frac{ft}{sec}[/tex]
Final velocity of the car V = 0
Acceleration of the car a = - 11 [tex]\frac{ft}{sec^{2} }[/tex]
First equation of motion,
V = u - at --------- (1)
Put all the values in equation 1, we get
0 = 40 - 11 t
⇒ t = [tex]\frac{40}{11}[/tex] = 3.636 [tex]\frac{ft}{sec^{2} }[/tex]
Now the distance traveled by the car before come into rest
[tex]S = u t + \frac{1}{2} a t^{2}[/tex]
Put all the values in the above equation we get,
S = 40 × 3.636 - [tex]\frac{1}{2}[/tex] × 11 × [tex]3.636^{2}[/tex]
S = 145.44 - 72.71
S = 72.72 ft
This is the distance the have traveled before come into the rest.
