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Size:  338.3 KB I got 0.3125 for part c but the book has different answer. Thanks.
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    You found that total T for the journey is 96 s , 3/4 of that total T is for the time taken by the time spent on constant velocity, so that would be 72 s, 96 - 72 = 24,
    We are now told that deceleration is twice as acceleration, so if we take that as ratios, it would be a ratio of 2:1 for deceleration:acceleration , therefore 24/3 gives us 8, so the time spent on deceleration would be 8s and time spent on acceleration would be 16s . Now we know that it took 8s to decelerate till rest, and we know it took 16s to accelerate to 15ms^-1 (thus deceleration is twice the magnitude of acceleration as it took half the time for it to do).
    Now we split it into two separate suvat equations, using v = u + at for each, we then get :
    For deceleration : 0 = 15 + 8a , rearrange to get a and we get a = - 1.875 (Negative coz its decelerating)
    For acceleration : 15 = 0 + 16a, rearrange to get a we get a = 0.9375

    From the results we can see that the magnitude of deceleration is twice that of acceleration. 1.875 = 2(0.9375)

    Hope that helps
 
 
 
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