M2 Kinematics - A-level Math

Tullia SG

A body moves in straight line so that at time, t (s), its acceleration is -0.2(24 + v) ms^-2, where v ms-1 is its velocity. When t = 0, the body is at the origin and its velocity is 36 ms-1.

a) Show that v = 60e^-0.2t - 24.

b) Given that, at the time t (s), the displacement of the body from the origin is x (m), find an expression for x in terms of t.

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a)

a = dV/dt = -0.2(24 + v)

1/(24 + v) dV/dt = -0.2

∫1/(24 + v) dV = ∫ -0.2 dt

ln (24 + v) / 1 = -0.2t + c

ln (24 + 36) = 0 + c

c = ln(60)

ln(24 + v) = -0.2t + ln(60)

-0.2t = -ln(60) + ln(24 + v)

-0.2t = ln(24 + v) / 60

e^-0.2t = 24 + v / 60

60e^-0.2t = 24 + v

v = 60e^-0.2t - 24

b)

x = ∫ v dt

x = ∫ (60e^-0.2t - 24) dt

x = 60e^-0.2t / -0.2 - 24t + c

x = -300e^-0.2t - 24t + c

0 = -300e^0 - (24 x 0) + c

c = 300

x = -300e^-0.2t - 24t + 300

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If someone could please endeavour to work through my answers and correct/advise me where appropriate then it would be very much appreciated.

According to the answers I've got in my booklet b) should be:

x = -12e^-0.2t - 24

Thank you in advance.

dreamer87 SG

The booklet typo. Obviously, the answer given is wrong as x = -12e^-0.2t - 24, dx/dt not equal to v.

wee_ws SG

Hi Tullia,

Your working can be improved in terms of clarity:

a)

Acceleration a = dV/dt = -0.2(24 + v)

By the method of variable separable, we have

∫1/(24 + v) dV = ∫ -0.2 dt

=> ln (24 + v) / 1 = -0.2t + c.

When t = 0, v = 36, so ln (24 + 36) = 0 + c

=> c = ln(60).

Now ln(24 + v) = -0.2t + ln(60)

=> -0.2t = -ln(60) + ln(24 + v)

=> -0.2t = ln ((24 + v) / 60)

=> e^(-0.2t) = (24 + v) / 60

=> 60e^(-0.2t) = 24 + v

=> v = 60e^(-0.2t) - 24 (shown).

b)

x = ∫ v dt

=> x = ∫ (60e^(-0.2t) - 24) dt

=> x = (60e^(-0.2t)) / (-0.2) - 24t + c

=> x = -300e^(-0.2t) - 24t + c.

When t = 0, x = 0, so 0 = -300e^0 - (24 x 0) + c

=> c = 300.

Now x = -300e^(-0.2t) - 24t + 300 is our required particular solution.

Thanks.

Cheers,
Wen Shih