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# M2 Question watch

1. Soz this is a long one, feel free to only do some of the question if it's a bit much but i cant seem to get anywhere with it.

Question:

A particle P moves along the x-axis passing through the origin O at time t=0. At any subsequent time t seconds, P is moving with a velocity of magnitude v m/s in the direction of x increasing where:

v = 2t^3 + 2t + 3

a) Find the acceleration of P when t = 3
b) Find the distance covered by P between t = 0 and t = 4.

A second particle Q leaves O when t = 1 with constant velocity of magnitude 10 m/s in the direction of the vector 3i - 4j, where i and j are unit vectors parallel to 0x and 0y respectively.

Find, as a vector in terms or i and j
c) the velocity of Q
d) the velocity of P relative to Q at the instant when t=1

Hence
e) find the magnitude of the velociy of P relative to Q when t = 1
f) find the angle between the relative velocity and the vector i at the instant

a) 14 m/s
b) 70 2/3 m
c) (6i - 8j) m/s
d) (i + 8j) m/s
e) root 65 m/s
f) 82.9 degrees
v = 2t^2 + 2t + 3

a) Find the acceleration of P when t = 3
dv/dt = a
v = 2t^2 + 2t + 3
dv/dt = 4t + 2
at t= 3
a = 14 ms-1

b) Find the distance covered by P between t = 0 and t = 4.
S = int v dt
v = 2t^2 + 2t + 3
S = 2/3 . t3 + t² + 3t

S = 2/3 . (4)3 + (4)² + 3(4) - 0
S = 70 2/3 m
3. (Original post by yazan_l)
dv/dt = a
v = 2t^3 + 2t + 3
dv/dt = 6t² + 2
at t= 3
a = 56 ms-1

S = int v dt
v = 2t^3 + 2t + 3
S = ½t4 + t² + 3t

S = ½(4)4 + (4)² + 3(4) - 0
S = 156 m
Yeah, that's what i did but they are NOT the answers

a) 14 ms^-2
b) 70 2/3 m
4. Any suggestions?
5. A particle P moves along the x-axis passing through the origin O at time t=0. At any subsequent time t seconds, P is moving with a velocity of magnitude v m/s in the direction of x increasing where:

v = 2t^2 + 2t + 3

a) Find the acceleration of P when t = 3

a = dv/dt
a = 4t + 2

t = 3
a = 4(3) + 2
a = 14m/s²

b) Find the distance covered by P between t = 0 and t = 4.
v = dx/dt
∫1 dx = ∫v dt
x = ∫v dt = 2t3/3 + 2t2/2 + 3t

At t = 0, x = 0
At t = 4
x = 2(4)3/3 + 2(4)2/2 + 3(4)
x = 42.667 + 16 + 12
x = 70 and 2/3m
6. (Original post by Widowmaker)
I'm not so sure though

a = 56m/s² seems too high
7. Thanks anyway...
8. according to the answers : v = 2t² + 2t + 3 not v = 2t³ + 2t + 3
9. now try to solve the other questions, if ur problem was that u werent able to get the book's answers
10. here are the answers for the rest of the question, they are in white if you wanted them:

c) the velocity of Q
10 (3i-4j)/sqrt[3²+4²] =
10 (3/5 i - 4/5j) =
6i - 8j

d) the velocity of P relative to Q at the instant when t=1
VP at t=1 = 2 + 2 + 3 = 7i
velocity of P relative to Q = VP-VQ
velocity of P relative to Q = 7i - (6i - 8j) = i + 8j

Hence
e) find the magnitude of the velociy of P relative to Q when t = 1
|i+8j| = sqrt [ 1 + 8²] = sqrt [65]

f) find the angle between the relative velocity and the vector i at the instant
[email protected] = 8/1
@ = 82.9o

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