# M3 past paper questionWatch

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Thread starter 14 years ago
#1
A particle moves in a straight line so that its distanec, s metres, from the origin after time t seconds is given by s= t + sin 2t.
Find expressions for
(a) the velocity v ms^-1
(b) the acceleration, fms^-2

hence show that

f^2 = 16 - 4(v-1)^2

I can do part a and b no problem i am just struguling with the "hence show" part any help would be great thanks.
0
14 years ago
#2
(Original post by droid)
A particle moves in a straight line so that its distanec, s metres, from the origin after time t seconds is given by s= t + sin 2t.
Find expressions for
(a) the velocity v ms^-1
(b) the acceleration, fms^-2

hence show that

f^2 = 16 - 4(v-1)^2

I can do part a and b no problem i am just struguling with the "hence show" part any help would be great thanks.
a. v=1+2cos2t
b. f=-4sin2t

f^2=g
g=16sin^2 2t
=16(1-cos^2 2t)
=16 - 16(cos^2 2t)
=16-4(4cos^2 2t)
=16-4(v-1)^2
0
14 years ago
#3
(Original post by droid)
A particle moves in a straight line so that its distance, s metres, from the origin after time t seconds, is given by: s = t + sin 2t.

Find expressions for:

(a) the velocity v ms^-1
(b) the acceleration, fms^-2

Hence show that
f^2 = 16 - 4(v - 1)^2
a.) v = ds/dt = 1 + 2cos2t
b.) f = dv/dt = -4sin2t

f^2 = (-4sin2t)^2 = 16sin^2(2t) = 16[1 - cos^2(2t)] = 16 - 16cos^2(2t) = 16 - 4[4cos^2(2t)] = 16 - 4[2cos(2t)]^2 = 16 - 4{[2cos(2t) + 1] - 1}^2 = 16 - 4(v - 1)^2

QED.

Nima
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