Turn on thread page Beta

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

    • Thread Starter
    Offline

    19
    ReputationRep:
    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?
    Offline

    13
    ReputationRep:
    (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 T_2 = 75 - (5+T). Since the deceleration begins after 5+T seconds. Then you should only need the equation formed from considering the area under the graph.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by SherlockHolmes)
    Yes, your value of T is correct. Alternatively, you can express your T2 in terms of T as T_2 = 75 - (5+T). Since the deceleration begins after 5+T 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
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 14, 2017
Poll
Could you cope without Wifi?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.