Answer:
d = t(V2+V1)/ 2
t = 2d/(V2+V1)
t = 2(1.2)/ (0+25)
t = 0.096S
Explanation:
The stopping distance is known to be 1.20m. the final velocity is 0 meters per second and the starting or initial velocity is 25 meters per second.
Answer: 0.08s
Explanation:
From the question:
Initial velocity (u) = 30m/s
Final velocity (v) = 0
Distance (s) = 1.2m
Calculating the acceleration of the car using the third equation of motion ;
v^2 = u^2 + 2as
a = acceleration, s= distance
0 = 30^2 + (2 × a × 1.2)
0 = 900 + (2.4 × a)
a = (- 900) ÷ 2.4
a = - 375 m/s^2
Therefore, time taken
v = u + at
Where, t = time taken
0 = 30 + (-375 × t)
0 = 30 - 375 × t
375 × t = 30
t = 30 ÷ 375
t = 0.08 seconds
