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urgent Year 1 Statistics and Mechanics help!!!! watch

1. I'm trying to answer a question, to do with Velocity-time graph sketching.

This is the question: A car is Initially travelling with a constant velocity of 15 m/s for
'T'seconds. It then decelerates at a constant rate for
T/2 (1/2 T) seconds reaching a velocity of 10 m/s. It then
immediately accelerates at a constant rate for 3T/2 seconds
reaching a velocity of 20 m/s

a) sketch a velocity-time graph to illustrate the motion (3 marks)

EXTRA

b) Given that the car travels a total distance of 1312.5 m over the journey described, find the value of T (4 marks)

Thank you so much!
2. (Original post by KatheO11)
I'm trying to answer a question, to do with Velocity-time graph sketching.

This is the question: A car is Initially travelling with a constant velocity of 15 m/s for
'T'seconds. It then decelerates at a constant rate for
T/2 (1/2 T) seconds reaching a velocity of 10 m/s. It then
immediately accelerates at a constant rate for 3T/2 seconds
reaching a velocity of 20 m/s

a) sketch a velocity-time graph to illustrate the motion (3 marks)

EXTRA

b) Given that the car travels a total distance of 1312.5 m over the journey described, find the value of T (4 marks)

Thank you so much!
Please tell us what you've done so far. Have you sketched the graph? What specifically are you having trouble with?
3. (Original post by Notnek)
Please tell us what you've done so far. Have you sketched the graph? What specifically are you having trouble with?
It's the graph sketching that is an issue, if it starting from 15, with T, how would I be able to draw 10 at T/2? How would I draw it?
4. (Original post by KatheO11)
It's the graph sketching that is an issue, if it starting from 15, with T, how would I be able to draw 10 at T/2? How would I draw it?
It travels constant speed for T seconds so you should mark T on the x-axis when it starts to decelerate.

Then it takes T/2 seconds for it to get to 10 m/s. This means that the total time at this point will be T + T/2 ( = 3/2 T) so mark this on your x-axis at the point where it is 10 m/s. Here's what the start of your graph will look like:

5. (Original post by Notnek)
It travels constant speed for T seconds so you should mark T on the x-axis when it starts to decelerate.

Then it takes T/2 seconds for it to get to 10 m/s. This means that the total time at this point will be T + T/2 ( = 3/2 T) so mark this on your x-axis at the point where it is 10 m/s. Here's what the start of your graph will look like:

Ok, I understand that. But how about 3T/2 reaching a velocity of 20 m/s?
6. (Original post by KatheO11)
Ok, I understand that. But how about 3T/2 reaching a velocity of 20 m/s?
At the end of my graph above it's been traveling for 3T/2 seconds. Then it accelerates for 3T/2 so at the end of the acceleration it will have been traveling for 3T/2 + 3T/2 = 3T seconds so mark "3T" on the x-axis at the point where it's 20 m/s.
7. (Original post by Notnek)
At the end of my graph above it's been traveling for 3T/2 seconds. Then it accelerates for 3T/2 so at the end of the acceleration it will have been traveling for 3T/2 + 3T/2 = 3T seconds so mark "3T" on the x-axis at the point where it's 20 m/s.
Thank you so much!
8. How do you answer to part b to this question?
9. (Original post by lolseriously?)
How do you answer to part b to this question?
When you've completed the graph sketch, you'll see that the three distinct phases of motion (constant, decelerating and accelerating) can be split, by dotted vertical lines so to speak, into three areas under the graph which together will have an area equal to 1312.5

Those three pieces are a rectangle and two trapezia, and you should be able to obtain expressions for their respective areas in terms of T.

For the first, the "length" of the rectangle is T and its "height" is 15, so the area is 15T.

For the second, the two parallel sides of the trapezium are 15 and 10 and the perpendicular distance between them is T/2, so by the usual formula, the area here is T/4 (15 + 10) = 25/4 T.

Finally for the third, the two parallel sides of the trapezium are 10 and 20 and the perp. dist. between them is 3T/2, so similarly to the above, we obtain an area of 22.5 T.

So in total we have (15 + 6.25 + 22.5) T = 43.75 T = 1312.5

from which we obtain T = 30.
10. In the suvat problems, remember this:
if you have 3 known values, you can always calculate the other two

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