Answer:
The time taken for the car to stop is 5.43 s.
The initial velocity of the car is 108.6 ft/s
Explanation:
The following data were obtained from the question:
Acceleration (a) = –20 ft/s² (since the car is coming to rest)
Distance travalled (s) = 295 ft
Final velocity (v) = 0 ft/s
Time taken (t) =?
Initial velocity (u) =?
Next, we shall determine the initial velocity of the car as shown below:
v² = u² + 2as
0² = u² + (2 × –20 × 295)
0 = u² + (–11800)
0 = u² – 11800
Collect like terms
0 + 11800 = u²
11800 = u²
Take the square root of both side
u = √11800
u = 108.6 ft/s
Therefore, the initial velocity of the car is 108.6 ft/s.
Finally, we shall determine the time taken for the car to stop as shown below:
Acceleration (a) = –20 ft/s² (since the car is coming to rest)
Final velocity (v) = 0 ft/s
Initial velocity (u) = 108.6 ft/s
Time taken (t) =?
v = u + at
0 = 108.6 + (–20 × t)
0 = 108.6 + (–20t)
0 = 108.6 – 20t
Collect like terms
0 – 108.6 = – 20t
– 108.6 = – 20t
Divide both side by –20
t = – 108.6 / –20
t = 5.43 s
Therefore, the time taken for the car to stop is 5.43 s.
