A car accelerates at a constant rate of change of velocity in 5 seconds from 10 m/s to 25 m/s on a highway. What is the acceleration and how far does the car go in this time-interval? Give a graphical solution.
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the change in v =(25-10)m/s
the change in v =(25-10)m/s=15m/s
the change in v =(25-10)m/s=15m/st=5s
the change in v =(25-10)m/s=15m/st=5sa=v÷t
the change in v =(25-10)m/s=15m/st=5sa=v÷t=15÷5
the change in v =(25-10)m/s=15m/st=5sa=v÷t=15÷5=+3m/s^2
the change in v =(25-10)m/s=15m/st=5sa=v÷t=15÷5=+3m/s^2distance travel= area of tapezium
the change in v =(25-10)m/s=15m/st=5sa=v÷t=15÷5=+3m/s^2distance travel= area of tapezium1÷2× (sum of prll sides )×h
the change in v =(25-10)m/s=15m/st=5sa=v÷t=15÷5=+3m/s^2distance travel= area of tapezium1÷2× (sum of prll sides )×h÷1÷2×(10+25)×5
the change in v =(25-10)m/s=15m/st=5sa=v÷t=15÷5=+3m/s^2distance travel= area of tapezium1÷2× (sum of prll sides )×h÷1÷2×(10+25)×562.5m is the answer
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