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M1 -- Calculating T from a speed-time graph. watch

1. Question:
An athlete runs along a straight road. She starts from rest and moves with constant acceleration for 5 seconds, reaching a speed of 8m/s. This speed is then maintained for T seconds. She then decelerates at a constant rate until she stops. She has run a total of 500m in 75(s).

b) Calculate the value of T

My issue:
So I drew the graph (I can't post it due to technical difficulties) and I have two unknowns, T and T2. I then presumed I had 2 simultaneous equations.

First equation -- T + T2 = 70 (added up the seconds, minused 5 from 75...)
Second equation -- 8T + 4T2 = 480 (area under the graph, minused 20 from 500...)
Third equation -- 4T + 4T2 = 280 (I multiplied the first equation by 4 so I cancel out T2)

I then did my working out and I got T = 50 seconds. Is this correct?
2. (Original post by Ze Witcher)
Question:
An athlete runs along a straight road. She starts from rest and moves with constant acceleration for 5 seconds, reaching a speed of 8m/s. This speed is then maintained for T seconds. She then decelerates at a constant rate until she stops. She has run a total of 500m in 75(s).

b) Calculate the value of T

My issue:
So I drew the graph (I can't post it due to technical difficulties) and I have two unknowns, T and T2. I then presumed I had 2 simultaneous equations.

First equation -- T + T2 = 70 (added up the seconds, minused 5 from 75...)
Second equation -- 8T + 4T2 = 480 (area under the graph, minused 20 from 500...)
Third equation -- 4T + 4T2 = 280 (I multiplied the first equation by 4 so I cancel out T2)

I then did my working out and I got T = 50 seconds. Is this correct?
Yes, your value of T is correct. Alternatively, you can express your T2 in terms of T as . Since the deceleration begins after seconds. Then you should only need the equation formed from considering the area under the graph.
3. (Original post by SherlockHolmes)
Yes, your value of T is correct. Alternatively, you can express your T2 in terms of T as . Since the deceleration begins after seconds. Then you should only need the equation formed from considering the area under the graph.
Thanks for that -- I tried to do that single equation but I thought I was wrong -- I made a silly mistake prior to that calculation which meant it didn't work but I see it now

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