Answer:
See explanation
Explanation:
We convert the speed of the car to ms-1
90 × 1000/3600 = 25 ms-1
a) from;
F= ma
a= F/m
a= 8400/1050 = 8 ms-2
b) from
v= u - at
v= 0
u = 25 ms-1
a= 8 ms-2
t= ?
t = u/a = 25/8 = 3.1 s
From;
v^2 = u^2 - 2as
v= 0 ms-1
u^2 = 2as
s = u^2/2a
s= (25)^2/2 × 8
s= 625/16
s = 39 m
A. The acceleration of the car when the brakes were applied is –8 m/s².
B. The time taken for the car to stop is 3.13 s.
C. The distance travelled by the car while the brakes were applied is 39.06 m.
A. Determination of the acceleration of car.
Force (F) = –8400 N (opposite direction)
Mass (m) = 1050 kg
Force = mass × acceleration
–8400 = 1050 × a
Divide both side by 1050
a = –8400 / 1050
NOTE: The negative sign indicates that the car is coming to rest.
B. Determination of the time taken for the car to stop.
Initial velocity (u) = 90 Km/h = (90 × 1000)/3600 = 25 m/s
Final velocity (v) = 0 m/s
Acceleration (a) = –8 m/s²
v = u + at
0 = 25 + (–8 × t)
0 = 25 – 8t
Collect like terms
0 – 25 = –8t
–25 = –8t
Divide both side by –8
t = –25 / –8
Therefore, the time taken for the car to stop is 3.13 s
C. Determination of the distance travelled by the car.
Initial velocity (u) = 25 m/s
Final velocity (v) = 0 m/s
Acceleration (a) = –8 m/s²
v² = u² + 2as
0² = 25² + (2 × –8 × s)
0 = 625 – 16s
Collect like terms
0 – 625 = –16
–625 = –165
Divide both side by –16
s = –625 / –16
Therefore, the distance travelled by the car is 39.06 m
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