1. ## kinematics thing again

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?

2. Hello furor celtica
Originally Posted by 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.

3. one thing that confused me is this thing
if v=u+at, then s=vt would make s=(u+at)t= ut +at^2
why is this different from the formula s=ut + 0.5at^2 which i was told to memorize? does this have to do with average velocity as opposed to end velocity?

4. you said
"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"
why?
i just tried to find the intersection of the curves representing the distance travelled by the two mobiles after 8 seconds, since this time is when the distance covered by the bike becomes a linear function. is this method incorrect?

5. Hello furor celtica
Originally Posted by furor celtica
one thing that confused me is this thing
if v=u+at, then s=vt would make s=(u+at)t= ut +at^2
why is this different from the formula s=ut + 0.5at^2 which i was told to memorize? does this have to do with average velocity as opposed to end velocity?
Yes, $\displaystyle s = vt$ applies only if $\displaystyle v$ is constant, or if you take the average velocity over a period of constant acceleration. If $\displaystyle u$ and $\displaystyle v$ are the initial and final velocities, then $\displaystyle s = \tfrac12(u+v)t$.
you said
"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"
why?
i just tried to find the intersection of the curves representing the distance travelled by the two mobiles after 8 seconds, since this time is when the distance covered by the bike becomes a linear function. is this method incorrect?
Think about it. At this point on the $\displaystyle s-t$ graph, the tangent to the curve is parallel to the straight line. This is where the vertical distance between them is greatest. And, on the ground, this is the point at which the car starts to overtake the motor-bike,