Motion along a straight line
A particle A starts to move from a fixed point O along a straight line. Its velocity in m s^-1 is given by v = 16 - 2t - 3(t^2) where t is the time in seconds after the particle has left O. Particle B moves in the same straight line, starting from O with a velocity of 6 m s^-1 at the instant the particle A passes throught point O. Particle B moves with an acceleration a = 4t - 8 . Find
i) the velocity of particle A when it passes through point O.
ii) the velocity of particle B when its displacement from particle A is at its maximum.
The velocity of A is a vector, that is it has a magnitude and a direction (forwards or backwards). The velocity is given by
`v = 16 - 2t - 3t^2`
The distance travelled by A is given by the integral of this with respect to t:
`d = 16t - t^2 - t^3`
A returns to O when the distance covered is again zero, ie when
`16t - t^2 - t^3 = 0` or
`-t(t^2 + t - 16) = 0`
`t=0` is a solution to this, because A starts at O. The other solution is obtained by solving the quadratic in brackets using the quadratic formula:
`t = (-1 pm sqrt(1 + 64))/2 = (-1 +sqrt(65))/2` (t must be positive remember)
i) The velocity of A at this time is
`v = 16 - (-1 +sqrt(65)) - 3/4(1 - 2sqrt(65) + 65) = -28.5` m/s
That is, 28.5 m/s 'backwards'.
ii) The velocity of particle B is given by the integral of its accleration. We are given that B starts at a velocity of 6 m/s from O when A passes through O (when t = `(-1 +sqrt(65))/2` s)
and that the acceleration of B is
`a = 4t - 8`
The velocity of B is then
`v = 2t^2 - 8t + c` where `c` is a constant
Now, at ` `the time when A passes through O and B starts to move
`v = (1-2sqrt(65)+65) - 4(-1 +sqrt(65)) + c = -3.31 + c `
We also know that at that time `v = 6` so that `c = 6 + 3.31 = 9.31`.
Therefore B's velocity at time t (measured from when A starts to move) is
`v = 2t^2 -8t + 9.31`
The distance B has travelled at time t is the integral of this
`d = (2/3)t^3 - 4t^2 + 9.31t`
The distance (or displacement) between A and B is greatest when `d_B - d_A ` is at its maximum, ie when
`(2/3)t^3 - 4t^2 + 9.31t - (16t - t^2 - t^3)` is maximised with respect to t. That is when
`(5/3)t^3 - 3t^2 - 6.69t` is maximised with respect to t.
To find this time, differentiate with respect to t and set to zero thus:
`5t^2 - 6t - 6.69 = 0`
`t = (6 pm sqrt(36 - 4(5)(-6.69)))/10 = (6 + sqrt(169.8))/10` (t must be positive)
At that time the velocity of B is
`v = 2/100(36+12sqrt(169.8) +169.8) - 8/10(6+sqrt(169.8)) + 9.31`
`= 1.3` m/s to 1 dp
i) -28.5 m/s to 1 dp
ii) 1.3 m/s to 1 dp
The question is a bit confusing as it asked that what is its velocity when the particle passes through the point. Does it mean that the question wants us to find its velocity when it passes through point O (comes to point O from another point) ? Or the question wants us to find its initial velocity? The answer is 16 m s^-1, that is when t=0 (initial velocity)