Question 1.12: The acceleration of a particle (in ms^-2) at time t seconds ...

The information given may be summarised as follows:

at t = 0, s = 0 and v = -2;

at time t, a = 6 – t.

i) \frac{dv}{dt} = a = 6  –  t

Integrating gives

v = 6t  –  \frac{1}{2}t^{2} + c.

When t = 0, v = -2

so    -2 = 0 – 0 + c

c = -2

At time t

v = 6t  –  \frac{1}{2}t^{2}  –  2.

ii)  \frac{ds}{dt} = v = 6t  –  \frac{1}{2}t^{2}  –  2

Integrating gives

s = 3t^{2}  –  \frac{1}{6}t^{3}  –  2t + k.

When t = 0, s = 0

so      0 = 0 – 0 – 0 + k

k = 0.

At time t

s = 3t^{2}  –  \frac{1}{6}t^{3}  –  2t.

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Notice that two different arbitrary constants (c and k) are necessary when you integrate twice. You could call them c_{1} and c_{2} if you wish.

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The three numbered equations can now be used to give more information about the motion in a similar way to the suvat equations. (The suvat equations only apply when the acceleration is constant.)

When   t = 6      v = 36 – 18 – 2 = 16                 from ②

When   t = 6      s = 108 – 36 – 12 = 60            from  ③

The particle has a velocity of +16 ms^{-1} and is at +60 m after 6 s.