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#1
1) A body moves in a straight line. At t seconds the acceleration of the body is a = 5t + 3. When t = 0 the velocity of the body is 4ms−1 and it’s displacement is 2m.
￼Find an expression for v and s in terms of t

3) Solve the following differential equation using the fact that when y = 7 then x = 0 dy/dx − y = 2x

Any help would be greatly appreciated
0
4 years ago
#2
Is this M1?
I hope this helps

1) I think what you need to do here is integrate.

To get from acceleration to velocity, you integrate a=5t+3 to get v= 5/2t^2 + 3t + c.
You know that when t=0 v=4 ,so c=4.

Now, to get to displacement you integrate v = 5/2t^2 + 3t + 4 to give you s = 5/6t^3 + 3/2t^2 + 4t + c
Since you know that when t=0 s=2, you can infer that c = 2

v= 5/2t^2 + 3t + 4
s = 5/6t^2 + 3/2t + 4t + 2

2) Do you have any more information on this question? I'm not sure I understand what it's asking.
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