(a) After 6.0 s, the car's velocity is
a*t = 2.4 * 6 = 14.4 m/s and it will have travelled
X = (a/2) t^2 = 86.4 m.
Three seconds of braking at -1.0 m/s^2 then reduces the velocity, by 3.0 m/s, from
14.4 to 11.4 m/s.
(b) Additional distance travelled while braking
= (avg. speed durng interval)*3.0 s
= 12.9 * 3 = 38.7 m
Add 38.7 m to 86.4 m for the total distance travelled during the 9.0 seconds.
A car starts from rest and travels for 6.0 s with a uniform acceleration of +2.4 m/s2. The driver then applies the brakes, causing a uniform acceleration of −1.0 m/s2. If the brakes are applied for 3.0 s, determine each of the following.
(a) How fast is the car going at the end of the braking period?
(b) How far has the car gone?
2 answers
4.65