# 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...

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.

mathsworkmusic | (Level 2) Educator

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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