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