Question

A particle P is moving along a straight line AB with its initial velocity 3 m/sec when

started from point A and with acceleration of 1.5 m/sec 2 . Another particle Q starts from

point A with velocity of 11.4 m/sec and acceleration of 1.2 m/sec 2 , 3 seconds after P has

left the point A. (a) determine the distance from point A when Q will overtake P, (b)

Further show that once Q will overtake P, Q will never be ahead of P by more than 9.6

m.

Answer 1

Let $t=$ time of motion of particle Q. Then the time of motion of particle P will be $t+3.$

The equation of motion

$$s=s_0+v_0t+\dfrac{at^2}{2}$$

Given

$$s_{0P}=0\ m,v_{0P}=3 \ m/s, a_P=1.5\ m/s^2$$

$$s_{0Q}=0\ m,v_{0P}=11.4 \ m/s, a_P=1.2\ m/s^2$$

$$s_P(t)=0+3(t+3)+\dfrac{1.5(t+3)^2}{2}$$

$$s_Q(t)=0+11.4t+\dfrac{1.2t^2}{2}$$

(a) Q will overtake P

$$3(t+3)+\dfrac{1.5(t+3)^2}{2}=11.4t+\dfrac{1.2t^2}{2}$$

$$3t+9+0.75t^2+4.5t+6.75-11.4t-0.6t^2=0$$

$$0.15t^2-3.9t+15.75=0$$

$$0.05t^2-1.3t+5.25=0$$

$$t^2-26t+105=0$$

$$(t-5)(t-21)=0$$

$t=5$

$$s_Q(5)=11.4(5)+\dfrac{1.2(5)^2}{2}=72$$

$$s_P(5)=3(5+3)+\dfrac{1.5(5+3)^2}{2}=72$$

$t=21$

$$s_Q(21)=11.4(21)+\dfrac{1.2(21)^2}{2}=504$$

$$s_P(21)=3(21+3)+\dfrac{1.5(21+3)^2}{2}=504$$

Q will overtake P at distance of 72 m from point A.

$$QP=11.4t+\dfrac{1.2t^2}{2}-(3(t+3)+\dfrac{1.5(t+3)^2}{2}),5\leq t\leq21$$

$$f(t)=QP=-0.15t^2+3.9t-15.75$$

$$t_{vertex}=-\dfrac{3.9}{2(-0.15)}=13$$

$$f(13)=QP_{t=13}=-0.15(13)^2+3.9(13)-15.75=9.6$$

The function $f(t)$ has the absolute maximum with value of $9.6$ for $5\leq t\leq21.$

Therefore once Q will overtake P, Q will never be ahead of P by more than 9.6 m.

P will overtake Q at $t=21$ sec.