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# M1 Kinematics help watch

1. Could someone please show me the method for part (c) of this question.:

14. A particle is initially at the origin. The particle has a constant acceleration of (2i-3j) m/s^2. Two seconds after it has left the origin, the velocity of the particle is (2i + 14j) m/s. The unit vectors i and j are perpendicular.

c) The speed of the particle is 10 m/s for the first time T seconds after it has left the origin. Find T.

The answer given in the book is 4 seconds but I'm not sure how to get to this. Help is appreciated
2. (Original post by vedderfan94)
Could someone please show me the method for part (c) of this question.:

14. A particle is initially at the origin. The particle has a constant acceleration of (2i-3j) m/s^2. Two seconds after it has left the origin, the velocity of the particle is (2i + 14j) m/s. The unit vectors i and j are perpendicular.

c) The speed of the particle is 10 m/s for the first time T seconds after it has left the origin. Find T.

The answer given in the book is 4 seconds but I'm not sure how to get to this. Help is appreciated
what info have you collected?
3. There is something missing from your information.

The particle can not have both constant acceleration (2i-3j) and constant speed (10m/s).
4. (Original post by msmith2512)
There is something missing from your information.

The particle can not have both constant acceleration (2i-3j) and constant speed (10m/s).
I didn't say anything about constant speed.
5. You need to find the initial velocity at the origin - which you can using v=u+at.

Then using it again to find a different velocity vector (v) which must have a magnitude of 10 units.

This new vector will have to be written using t (your unknown) but by equating its magnitude to 10 you will get a quadratic in t which you can solve.
6. (Original post by baphomet)
You need to find the initial velocity at the origin - which you can using v=u+at.

Then using it again to find a different velocity vector (v) which must have a magnitude of 10 units.

This new vector will have to be written using t (your unknown) but by equating its magnitude to 10 you will get a quadratic in t which you can solve.
The initial velocity is (-2i + 20j) m/s (that was part (a)). I used v = u + at like you said to get:

v = u + at
10 = (-2i + 20j) + (2i - 3j)t

What do I do now?
7. u = (-2i + 20j)
a = (2i - 3j)

For the speed of particle to equal 10 then i^2 + j^2 = 10^2

i^2 + j^2 = 100

The i component at time t = -2 + 2t
The j component at time t = 20 -3t

Therefore (-2+2t)^2 + (20-3t)^2 = 100

Expand, factorise and find t.

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Updated: September 27, 2011
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