Urgent Physics help?

Question

The rear of a bicycle passes a point O on a road with a velocity 4ms-^1 and an acceleration of 2ms^1.4 seconds later the front of a car passes O with a velocity of 2ms-^1 and an acceleleration of 4ms-^1.How far from O does the front of the car meet the rear of the bicycle? Please help so may of times q's like these come up nd i cant do em

Answer 1

Use x(t)=xi+vxi*t+.5*ax*t^2

For the rear of the bike,
xb(t)=4*t+.5*2*t^2

for the front of the car
xc(t)=2*(t-1.4)+.5*4*(t-1.4)^2

the two intersect when
xb(t)=xc(t)
so
4*t+.5*2*t^2=
2*(t-1.4)+.5*4*(t-1.4)^2
solve for t

7.44 seconds

note, there is also a root at 0.16 seconds. Ignore this since the equation for the car is only valid for t>=1.4

j