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# Equations of Motion Questions Help watch

1. Hi guys. Started doing this sheet for homework and been ok till these questions. I know initially they are very easy questions but what i'm stuck on is working out the right formula given what i get in the question. So if you can give me any hints for these ill be very greatful and i'm not expecting the answers to them, just some hints and tips. Thanks

1. A drag car accelerates from rest for 10s at 5ms(2) It then travels at a constant speed for 20s. The braking parachute is deployed and the drag car takes a further 25s to come to rest.
a) What is the speed at the end of the acceleration phase?
b) What is the total distance covered by the drag car from start to finish?

2. A car accelerates from rest for 10s and reaches a speed of 15m/s. It maintains this speed for 15s and then decelerates at 2ms(2) until it comes to rest.
a) What is the initial acceleration?
b) How long does it decelerate?
c) How ar does the car travel?
d) What is the average speed during the journey.

If say the answer to part a) determines the answers to the next questions, then you don't have to give me any hints to the next ones if it is relatively easy to find out the right formula given that I am struggling with these anyway at the moment.

Thank you! x
2. Start by writing:

a = ..., u = ..., v = ..., t = ...

And say you need to find s, add s = ?

Now use appropriate formulas.
3. 1a) u=0 a=5 t=10, you need to find v, so use v = u + at.
1b) Probably be easier for you to understand if you drew a graph of the whole journey, velocity on the y axis, time on the x axis then work out the area under the graph which is the total distance.
4. Here are (some) equations of motion:

i. v = v0 + at
ii. x = x0 + v0t + 1/2at^2 (This is the integral of equation i with respect to time)
iii. v^2 = v0^2 + 2a(x-x0)

Be aware that the acceleration in these equations is assumed to be constant. If the acceleration is changing, then you generally need to use calculus. This problem isn't that hard though.

x0 is initial displacement, x is final displacement, v is final velocity, v0 is initial velocity, a is acceleration, and t is the change in time (normally shown as delta-t)

So for 1:
a) v0 = 0, a = 5m/s^2, and t = 10s
So v = 5*10 m/s = 50 m/s

b) Take this in parts. There are three phases to the car's acceleration, and you should calculate the distances travelled in each phase - one phase for the first 10 seconds (equation ii, x = 1/2*5*10^2 m = 250m), one phase for the middle 20s at constant velocity (use equation i, x = 250m + 50m/s * 20s = 1250m), and one phase for the final deacceleration, which is constant and at -2m/s^2 (by equation ii, x = 1250m + 50*25 - (1/2)*(-2)*25^2 = 2500m - 625m = 1875m. 1875m is the final distance travelled.

Edit: The way I've done this is to calculate the displacement over the first 10 seconds, then the displacement for the next 20 seconds plus the displacement for the first 10 seconds, and then finally the displacement during the last 25 seconds added to the previous. You could just have had an initial displacement of 0 for each period of time and added the three together. I hope this should deconfuzzle and confusing steps I have made.

Hope this helps,
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