Hello furor celtica Quote:

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**furor celtica** 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?

I attach the $\displaystyle v-t$ and $\displaystyle s-t$ graphs.

Suppose the two are side by side $\displaystyle t$ seconds after the start.

Using the $\displaystyle v-t$ graph, the distance travelled is represented by the area under the graph. The area under each graph is the area of a triangle plus the area of a rectangle. For the motor-bike, the length of the rectangle is $\displaystyle (t-8)$ and for the car it's $\displaystyle (t-20)$.

So, equating the two areas, we get: $\displaystyle \tfrac12.8.20+20(t-8)=\tfrac12.20.30+30(t-20)$

$\displaystyle \Rightarrow 80+20t-160 = 300+30t-600$

$\displaystyle \Rightarrow 220 = 10t$

$\displaystyle \Rightarrow t = 22$

So they're level again after $\displaystyle 22$ sec.

We can see from the s-t graph that the greatest distance apart will occur when the car's velocity (the gradient of the graph) is equal to the motor-bike's velocity; i.e. $\displaystyle 20$ m per sec. The car will take $\displaystyle \frac{20}{1.5}=\frac{40}{3}$ sec to accelerate to this speed.

In this time, the motor-bike has covered $\displaystyle 80 + 20\left(\frac{40}{3}-8\right) =\frac{560}{3}$ m; and the car has covered $\displaystyle \frac12.\frac{40}{3}.20 = \frac{400}{3}$ m. So the greatest distance apart $\displaystyle = \frac{560-400}{3}=\frac{160}{3}$ m.

Grandad