d1 = 0.5a*t^2 = 0.6*5.3^2 = 16.85 m.
V1 = a*t = 1.2m/s^2 * 5.3s = 6.36 m/s.
a. V = Vo + a.t
V = 6.36m/s - 2.3m/s^2*1.60s = 2.68 m/s.
b. V^2 = V0^2 + 2a*d
d2 = V^2-Vo^2)/2a
d2 = (2.68^2-6.36^2)/-4.6 = 7.23 m.
D = d1 + d2 = 16.85 + 7.23 = 24.1 m.
A car starts from rest and travels for 5.3 s with a uniform acceleration of +1.2 m/s2. The driver then applies the brakes, causing a uniform acceleration of -2.3 m/s2. The breaks are applied for 1.60 s.
How fast is the car going at the end of the braking period?
How far has the car gone from its start?
1 answer
1 answer