v = Vo + at
s = Vo*t + 1/2 at^2
so,
v = 0 + 3(10) + 2(-4) = 22 m/s
s = (0*10 + 1/2 * 3 * 10^2) + (30*2 - 1/2 * 4 * 2^2) = 202 m
A car starts from rest and travels for 10 seconds with a constant acceleration of 3.0 m s^-2. The driver then applies the brakes causing a negative acceleration of -4.0 m s^-2. Assuming that the brakes are applied for 2.0 seconds, calculate,
a) the velocity of the car at the end of braking
b) the distance travelled by the car at the end of braking
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2 answers