The Student Room Group

Physics A-Level (Modelling Physics OCR Spec query)

How do i answer part B?

16 (a) Two cars, A and B, are travelling clockwise at constant speeds around the track shown in Fig. 16.1.
The track consists of two straight parallel sections each of length 200 m, the ends
being joined by semi-circular sections of diameter 80 m.
The speed of A is 20 m s
–1
and that of B is 23 m s
–1
.
Fig. 16.1
(i) Calculate the time for A to complete one lap of the track.
time for one lap = ……….………….. s [2]
(ii) Starting from the positions shown in Fig. 16.1 determine the shorter of the two distances along
the track between A and B, immediately after A has completed one lap
Im assuming you mean ii) - you haven't posted part b.
So in part a i, you worked out the time as 33s
so B has travelled ~750m
One lap is ~650m so its travelled around 100m to the right
The shorter distance along the track is half the distance of the track minus the distance b has travelled
so about 325-100 = ~225m.
I didnt use exact values but you get the idea.
Hope that helps :smile:
yh i meant ii, thanks for that, could you help me with part B ii tho aswell, i know we need to use s=ut + 1/2at^2 but im confused as to how the mark scheme in order to find the distance for B used that equation with two different t values, this is what they wrote: Distance moved by B = (1/2x1.5x162 ) + 24(t 16), how can you use two different t values?



(b) Cars A and B are now on a straight road with car A moving at 22 m s–1 and car B at rest. As car A passes car B, car B accelerates from rest in the same direction at 1.5 m s –2 for 16 s. It then moves with constant velocity. Fig. 16.2 shows the graph of velocity against time for car A. The time t = 0 is taken when the cars are alongside. Fig. 16.2 (i) Sketch the graph of velocity against time for car B on Fig. 16.2. [2]

(ii) Determine the time taken for car B to be alongside car A.
(edited 1 month ago)
basically time starts when t=0, 16s later b has accelerated to its constant velocity of 24ms^-1 so now t=16s.
From this point b will travel further distance at its constant velocity for some time t - but this variable t includes the starting 16s, so the time that b has spent going at constant velocity is (t-16)s
So b distance is = (1/2x1.5x162 ) + 24(t 16)
A has travelled at constant velocity for the whole duration so its distance is just 22t
Hope that makes sense :smile: and let me know if you need any further help.

and btw there is no different t values, the variable t always represents the same amount of time, we can just add/subtract some time from t to get the time that we want.
Original post by mosaurlodon
Im assuming you mean ii) - you haven't posted part b.
So in part a i, you worked out the time as 33s
so B has travelled ~750m
One lap is ~650m so its travelled around 100m to the right
The shorter distance along the track is half the distance of the track minus the distance b has travelled
so about 325-100 = ~225m.
I didnt use exact values but you get the idea.
Hope that helps :smile:

the part about the "shorter of the two distances is tripping me up". I dont understand what its asking.
Original post by tbkoli88
the part about the "shorter of the two distances is tripping me up". I dont understand what its asking.

There are two distances between A and B, one in the clockwise direction and one in the anticlockwise direction, because they are not spaced equally apart one going in one direction is shorter than the other.

Quick Reply

Latest