kinematics thing again

[red and white kop!]

ok hang onto your seats cos this is kinda long

a motorbike and a car are waiting side by side at traffic lights.
when the lights turn green, the motorbike accelerates at 2.5 ms^-2 up to a top speed of 20ms^-1 and the car accelerates at 1.5 ms^-2 up to a top speed of 30ms^-1. both then continue to move at a constant speed. draw (t,v) graphs for each vehicle, using the same axes, and sketch the (t,s) graphs.
A. after what time will the motorbike and the car again be side by side?
B. what is the greatest distance that the motorbike is in front of the car?

for A I followed the (t,s) graph and tried to find the intersection of the two curves after 8 seconds, point at which the bike ceases to accelerate. i found 24s but my book says 22s. can somebody check this out? i did my calculations over and over.

for B i avoided using the (t,v) gaph but turned up with distance after 8s = 64m. this didnt fit in and the result was wrong, so i used geometric formulae on the (t,v) graph and got distance after 8s = 32m and the correct answer to the B (53.3 m). can somebody go over the same calculations and tell me where things went wrong?

 

[wjm11]

a motorbike and a car are waiting side by side at traffic lights.
when the lights turn green, the motorbike accelerates at 2.5 ms^-2 up to a top speed of 20ms^-1 and the car accelerates at 1.5 ms^-2 up to a top speed of 30ms^-1. both then continue to move at a constant speed. draw (t,v) graphs for each vehicle, using the same axes, and sketch the (t,s) graphs.
A. after what time will the motorbike and the car again be side by side?
B. what is the greatest distance that the motorbike is in front of the car?

for A I followed the (t,s) graph and tried to find the intersection of the two curves after 8 seconds, point at which the bike ceases to accelerate. i found 24s but my book says 22s. can somebody check this out? i did my calculations over and over.

You didn’t post your calculations, so I cannot say where you went wrong. You mentioned something about doing your calculations at 8s, though, so I suspect you may have made a wrong assumption about one of the vehicles (probably the car) at that point.

The car and bike both start with zero velocity and from the same spot, so the displacement equation simplifies to

x(t) = .5at^2.

And the velocity equation is

v(t) = at

The bike a = 2.5 and max v = 20:

t = 20/2.5 = 8s

The car a = 1.5 and max v = 30:

t = 30/1.5 = 20s

Draw the displacement curve, x(t) vs t, for both vehicles on the same graph. The bike will be a parabola for the first 8s and then become a straight line with slope 20 m/s. The car will be a different parabola for 20s and then become a straight line with slope 30 m/s.

You need to examine three different regions to determine maximum distance between the vehicles to obtain max distance between them: 0-8s, 8-20s, and greater than 20s.

Hope that helps.