M2 review exercise

A particle P moves from rest at a point O at time t seconds along a straight line. at any subsequent time t seconds the acceleration of P is proportional to (7-t^2) m/s/s and the deisplacement of P from O is s metres. The speed of P is 6m/s when t=3.
show that:

s=(1/24)t^2(42-t^2)

hmm. just cant seem to get this!!

Also,

a particle P moves in a straight line in such a way that at time t secs, its velocity is given by
v=12t-3t^2 , ( t is greater or equal to ), but less than or equal to 5)
v=-375t^-2, (t is greater than 5)

calculate the displacement of P from O when
a) t=5
b)t=6

ans a) 25 b)12.5

i got a, but cant get b.

Reply 1

Christophicus

a = k(7 - t²) = 7k - kt²
v = ∫a dt + c
v = ∫7k - kt² dt
v = 7kt - kt³/3

v = 6, t = 3
6 = 7k(3) - k(3)³/3
6 = 21k - 9k
12k = 6
k = 6/12 = ½

v = 7t/2 - t³/6
s = ∫v dt
s = ∫7t/2 - t³/6 dt
s = 7t²/4 - t⁴/24
s = 42t²/24 - t⁴/24
s = (1/24)t²(42 - t²)

a particle P moves in a straight line in such a way that at time t secs, its velocity is given by
v=12t-3t^2 , ( t is greater or equal to ), but less than or equal to 5)
v=-375t^-2, (t is greater than 5)

calculate the displacement of P from O when
a) t=5
b)t=6

a)
v = 12t - 3t²
s = ∫v dt
s = ∫12t - 3t²
s = 12t²/2 - 3t³/3
s = 6t² - t³

At t = 5
s = 6(5)² - 5³
s = 150 - 125
s = 25m

b)
v = -375t^-2
s = ∫v dt + c
s = -375∫t^-2 dt + c
s = -375 x t^-2+1/(-2+1) + c
s = -375 x t^-1/-1 + c
s = 375/t + c

At t = 5, s = 25
25 = 375/5 + c
c = 25 - 75
c = -50

s = 375/t - 50

At t = 6
s = 375/6 - 50
s = 62.5 - 50
s = 12.5m

Reply 2

realicetic