I got 0.3125 for part c but the book has different answer. Thanks.
M1 last minute help please ! Watch
- Thread Starter
- 08-06-2016 00:23
- 08-06-2016 07:09
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