Two particles, P and Q, leave a point O at the same time and travel in the same direction along the same straight line. Particle P starts with a velocity of 9 m/s and moves with a constant acceleration of 1.5m/s2. Particle Q starts from rest and moves with an acceleration of a m/s2, where a = 1 + t/2 and t seconds is the time since leaving O. Find
(i) the velocity of each particle in terms of t,
(ii) the distance travelled by each particle in terms of t,
Hence find
(iii) the distance from O at which Q collides with P,
(iv) the speed of each particle at the point of collision.
Ahh…they rather naughty…gave the acceleration function. Hence, to get the velocity and displacement functions, we will need to integrate and find the constants of integration. OK, so let’s start…
(i) The velocity function for P is quite straight forward… vP = 9 + 1.5t
and for Q it is:
When t = 0, v = 0 so c = 0
Hence, vQ = t + t2/4.
(ii) The distance functions, denoted by sP and sQ , are obtained by integrating the velocity functions.
sP = 9t + 0.75t2 + c
sQ = t2/2 + t3/12 + c
When t = 0, s = 0, so c = 0
Hence, sP = 9t + 0.75t2 and sQ = t2/2 + t3/12
(iii) When both particles collide, it means that they will both have travelled the same distance. Hence, we will find the value of t when this happens and then find the distance travelled itself. So, we equate both the distance functions together and solve for t.
9t + 0.75t2 = t2/2 + t3/12
Multiplying throughout by 12 gives
108t + 9t2 = 6t2 + t3
t3 – 3t2 – 108t = 0
t2 – 3t – 108 = 0
(t – 12)(t + 9) = 0
t = 12 or t = -9
Hence, the particles will collide at t = 12s
The distance travelled is 9(12) + 0.75(12)2 = 216m
Hence, they are 216m from O when they collide.
(iv) to find the speed of each particle at each point of collision, we substitute t = 12s into the velocity functions.
Hence, P will be travelling at 9 + 1.5(12) = 27m/s and
Q will be travelling at 12 + (12)2/4 = 48m/s.
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