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Position vector of particle
By Anonymous on Sunday, May 27, 2001 - 10:40 pm:

A particle has position vector i+2j initially and is moving with speed 10ms-1 in the direction 3i-4j. Find its position vector when t=3 seconds and the distance it has travelled in those 3 seconds.

I appreciate the help guys

By Andrew Hodges (P4403) on Monday, May 28, 2001 - 02:25 pm:

If its speed is 10m/s in the direction 3i - 4j, you can find its velocity by considering the modulus of the direction {=(3×3 + -4×-4)0.5=5}. The speed is 10, which is 2×5, so the velocity vector is 2(3i-4j) = 6i-8j.

Remember that displacement (its position) is the integral of velocity => at time t, so the position vector is given by 6ti - 8tj + C, where C is the constant of integration. At t=0, we get 6×0i - 8×0j + i + 2j, which implies that C = i + 2j. Thus, the position vector at time 't' is..
(6t+1)i - (8t - 2)j, so at t=3, the position vector will be 19i - 22j.

The distance from i + 2j will be
mod(19i -22j-i-2j)
=mod(18i-24j) = 30m

By The Editor:

Or, since the speed is constant, distance = speed × time = 30m.