M2 Projectiles

Q1) A particle moves in a straight line such that at time t its displacement x from a fixed point O in that line is given by x = 3t^3 - 2t^2. Find (a) the times when it is at rest and (b) its distances from O when it is at rest

Answer: (a)t=0 and 4/9s (b)0m and 0.132m

Q2) A particle P moves in a straight line such that at time ts its acceleration is given by a= 2t - 2. It passes through a point O on the line when t=0 with a velocity of 1m/s. Find (a) the velocity of P when t=2 (b) the displacement of P from O when t=2 (c) the time when P is at rest (d) the distance of P from O when it is at rest.

Answer: (a) 1m/s
(b) 2/3m
(c) 1s
(d) 1/3m

Q3) A particle moves along the x-axis with velocity at time t given by v = 5t^2 + 2t. Find (a) the distance it moves in the 2nd second (b) the distance it moves in the 4th second.

Answer: (a) 14 2/3m
(b) 68 2/3m

Cheers.

Reply 1

JFN

Are you from XT?

Q1) A particle moves in a straight line such that at time t its displacement x from a fixed point O in that line is given by x = 3t^3 - 2t^2. Find (a) the times when it is at rest

x = 3t^3 - 2t^2
Differentiating:
dx/dt = v = 9t^2 - 4t
The particle is at rest at 9t^2 - 4t = 0
t(9t-4)=0
The particle is at rest when t = 0 and when t = 4/9.

(b) its distance from O when it is at rest

You need to find the distance at t=0 and t=4/9.
x = 3t^3 - 2t^2
at t=0, x=0m
at t=4/9, x=0.132m

Question Completed.

Reply 3

Gaz031

15

Q2) A particle P moves in a straight line such that at time t its acceleration is given by a= 2t - 2. It passes through a point O on the line when t=0 with a velocity of 1m/s. Find (a) the velocity of P when t=2

a = 2t-2
Integrating:
v = t^2 - 2t + c
You are given that t = 0, v = 1, x= 0
Hence: 1 = c
Hence:
v = t^2 - 2t + 1
When t = 2:
v = 4 - 4 + 1 = 1m/s

(b) the displacement of P from O when t=2

You now know that:
v = t^2 - 2t + 1
Integrating:
x = (1/3)t^3 - t^2 + t + c
As t = 0, v = 1, x = 0
0 = c
x = (1/3)t^3 - t^2 + t
At t = 2:
x = (1/3)8 - 4 + 2
x = 8/3 - 2
x = 8/3 - 6/3
x = (2/3)m

(c) the time when P is at rest

At rest when v = t^2 - 2t + 1 = 0
v = t^2 - 2t + 1 = 0 = (t-1)^2 = 0
Hence, t = 1 when the particle is at rest.

(d) the distance of P from O when it is at rest.

x = (1/3)t^3 - t^2 + t
at t = 0, x = (1/3) - 1 + 1
x = (1/3)m

Question complete.

Reply 4

Gaz031

15

Q3) A particle moves along the x-axis with velocity at time t given by v = 5t^2 + 2t. Find (a) the distance it moves in the 2nd second

v = 5t^2 + 2t
Integrating:
x = (5/3)t^3 + t^2 + c
at t = 1 second, x = 8/3 + c
at t = 2 seconds, x =52/3 + c
Difference = 14 2/3 m

(b) the distance it moves in the 4th second.

at t = 3 sec, x = 54 + c
at t = 4 sec, x = 368/3 + c
Difference = 68 2/3

Hope the solutions helped. Feel free to ask if you need anything explaining in more detail.

Remember:
x = x
dx/dt = v
d^2x/dt^2 = dv/dt = a.

Reply 5

Barça_Fan

OP

JFN

Are you from XT?

Yep

Who are you on XT?

By the way, thanks guys.

M2 Projectiles

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