Seemingly simple kinematics question

1. Sep 26, 2006

macmav

This question appears to be very simple at first but has stumped me for hours now:

we are given 2 accelerations
+7.5m/s^2 and -10m/s^
a car brakes and accelerates at these rates
over 105m, what is the maximum velocity the car will reach, starting and stopping at rest?

Originally I was thinking to use one of the three constant acceleration equations but I always end up going nowhere (0=0, 7.5t=10t) etc. The closest I've gotten was plotting a velocity-time graph but I can think of no way to find the constant to be added onto the line -10t to make the area under it and 7.5t equal 105m. I believe I just have to use the formula v^2 = vi^2 + 2(s-si) but every time I try I'm stuck!

If some1 can point me in the right direction itd be of awesomeness! Thank you for your time.

2. Sep 26, 2006

Staff: Mentor

That's the equation I would use. Break the motion into two segments: Speeding up (a1 = 7.5 m/s^2) and slowing down (a2 = -10 m/s^2). Don't forget that the distance traveled in each segment must add up to the total distance given.

3. Sep 26, 2006

neutrino

You're right about using the formula $$v^2 - u^2 = 2as$$. Write down two equations, one for the part where the car accelarates, and an other when the car decelarates. Add the two equations. (Make sure you get the signs right)

4. Sep 26, 2006

andrevdh

The maximum speed will be the speed at the end of the acceleration, just before the driver applies the brakes. Once the brakes is applied the speed of the car will start to decrease - that is the whole idea behind braking!

5. Sep 16, 2010

notnimdab2009

hi , i thought this problem is to be solved by maxima/minima of calculus but following above suggestions i arrived at the following,

1) The final velocity when the car is accelerating is the initial velocity when it is decelerating (brakes applied).
2) So i used the formula suggested above and found the vmax to be 30 m/sec and the distance/displacement achieved is 60 meters from starting pt and t=4 sec (applying the brake at t=4 sec).
3) i just tried to see what happens when the driver applied the brakes at t=5 secs, i got vmax of 37.5 but using this as initial velocity on the second part i found that final velocity is not zero, (34.37 m/sec).

I dont know the principle behind on why such simple equation for like this will yield the max velocity. Thanks to all.