Answer:
–2.25 m/s²
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 15 m/s
Distance travelled (s) = 50 m
Final velocity (v) = 0 m/s
Deceleration (a) =?
v² = u² + 2as
0² = 15² + (2 × a × 50)
0 = 225 + 100a
Collect like terms
0 – 225 = 100a
– 225 = 100a
Divide both side by 100
a = –225/100
a = –2.25 m/s²
Thus, the deceleration of the vehicle is –2.25 m/s²
Answer:
The car would travel after applying brakes is, d = 14.53 m
Explanation:
Given that,
The time taken to apply brakes fully is, t = 0.5 s
The velocity of the car, v = 29.06 m/s
The distance traveled by the car in 0.5 s, d = ?
The relation between the velocity, displacement, and time is given by the formula
d = v x t m
Substituting the values in the above equation,
d = 29.06 m/s x 0.5 s
= 14.53 m
Therefore, the car would travel after applying brakes is, d = 14.53 m
