Distance problem

Question

Firework A travels 220 fps. Firework B travels 200 fps. Firework B launches .25 seconds before firework A. How many seconds after firework B launches will they both explode?

Answer 1

Hey Natalie,

 

This is a "catch up" problem.  Fireworks A and B start from the same origin at different times, with Firework B being slower than A, but getting a head start.  Firework A then catches up and they explode on impact.

 

Begin by realizing that the distance traveled by both is the same, x.

For constant velocity, we have v=x/t, solve for t=x/v

Then

Let t_B=time from launch of B (at t=0) to explosion of B

Let t_A=time from launch of A (at t=0.25) to explosion of A

Then since they collide and explode t_B-0.25=t_A

Note, we are asked above to find t_B, so drop the B subscript 

We now have t and t-0.25 for Firework A and B times respectively.

 

For constant velocity, v=x/t ... or  x=vt

Since x is same for A and B, then can set ... v_At_A=v_Bt_B

                                                      220(t-0.25)=200(t)  

                                                                      t=220(0.25)/20

                                                                      t=2.75 sec

 

Check B goes 200 fps(2.75 sec)=550 ft

          A goes 220 fps(2.50 sec)=550 ft   YES!!