A car accelerates from rest with a constant positive acceleration of ax = +2 m/ss,?

Question

A car accelerates from rest with a constant positive acceleration of ax = +2 m/ss,
then brakes with a constant negative acceleration ax = −4 m/ss, coming to a stop after a
total time interval of 30 s has elapsed. What are the distance traveled and the maximum
velocity reached by the car? (b) What are the car’s average velocity and average acceleration
for this interval?

Answer 1

Let the amount of time under acceleration be t_1 and the time under deceleration be t_2. Since the maximum velocity of the car is reached at 2*t_1 and is back to 0 after -4*t_2, and the total time is 30 seconds, it must be that
t_1 + t_2 = 30
2*t_1 - 4*t_2 = 0
Knowing these will get you the maximum velocity, distance traveled, and the average velocity.


Doug

Answer 2

In general,
x = (a * t^2) / 2
V = a * t

If a1=2 and a2=-4, then a2=2*a1

also, t1 + t2 = 30 seconds, so t2 = 30 - t1

Total distance traveled

x = (a1 * t1^2 + a2 * t2^2 ) / 2 [remember that a2 is Negative, so this is a subtraction]

substitute for a2 and t2 as shown above, and solve for x.

Now you know the total distance traveled.

Then,

Vmax = a1 * t1

Vavg = x / 30