The acceleration of a particle (in ms^-2) at time t seconds is given by a = 6 - t . The particle is initially at the origin with velocity -2ms^-1. Find an expression for i) the velocity of the particle after t s ii) the position of the particle after t s. Hence find the velocity and position 6 s

Question

The acceleration of a particle (in [latex]ms^{-2}[/latex]) at time t seconds is given by

a = 6 - t .

The particle is initially at the origin with velocity -2 [latex]ms^{-1}[/latex]. Find an expression for

i) the velocity of the particle after t s
ii) the position of the particle after t s.

Hence find the velocity and position 6 s later.

Accepted Answer

The information given may be summarised as follows:

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

at time t, a = 6 - t.

i) [latex]\frac{dv}{dt} = a = 6 - t[/latex]

Integrating gives

[latex]v = 6t - \frac{1}{2}t^{2} + c.[/latex]

When t = 0, v = -2

so -2 = 0 - 0 + c

c = -2

At time t

[latex]v = 6t - \frac{1}{2}t^{2} - 2.[/latex]

ii) [latex]\frac{ds}{dt} = v = 6t - \frac{1}{2}t^{2} - 2[/latex]

Integrating gives

[latex]s = 3t^{2} - \frac{1}{6}t^{3} - 2t + k.[/latex]

When t = 0, s = 0

so 0 = 0 - 0 - 0 + k

k = 0.

At time t

[latex]s = 3t^{2} - \frac{1}{6}t^{3} - 2t.[/latex]

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Notice that two different arbitrary constants (c and k) are necessary when you integrate twice. You could call them [latex]c_{1}[/latex] and [latex]c_{2}[/latex] 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 [latex]ms^{-1}[/latex] and is at +60 m after 6 s.

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

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