Systems of Linear Equations and Inequalities >> Solving Systems of Equations by Substitution

Question

Pyrotechnician plans for two fireworks to explode together at the same height in the air. They travel at speeds shown below. Firework B is launched 0.25 s before Firework A. How many seconds after Firework B launches will both fireworks​ explode?

firework a = 340

firework b = 320

Both fireworks will explode [enter your response here] seconds after Firework B launches.

Answer 1

Pyrotechnician plans for two fireworks to explode together at the same height in the air. They travel at speeds shown below. Firework B is launched 0.25 s before Firework A. How many seconds after Firework B launches will both fireworks​ explode?

firework a = 340

firework b = 320

Both fireworks will explode [enter your response here] seconds after Firework B launches.

So with this problem we have 2 unknowns - we don't know the time the fireworks are in the air and we don't know the total distance they travel. because we have 2 unknowns, we'll need 2 equations to solve for the variables. We know that speed x time is equal to distance, so we'll set up 2 equations (one for firework a and one for firework b) to solve for this.

Let's start with firework B, as this is firework that the time will be based off of as per the question.

Firework B goes at a speed of 320, so the Distance for b, Db, for firework B would be:

Db = 320* t,

at t=0, Db=0

at t=1, Db=320

at t=2, Db=640

etc....

where t is the time in seconds that the firework is traveling

for Firework A, we know that it travels at 340, so we can say:

Da = 340*t,

BUT! we must remember that firework A leave 0.25s after firework B, so we need to subtract out the distance lost by that 0.25 seconds.

Da = 340*t - 0.25*340 = 340*t - 85

so we know that in this case, when t=0.25, Da will be 0

so with our 2 equations, we have:

Da = 340*t - 85

Db = 320* t

we know we want the distances to be equal, so Da = Db, so we will substitute Da for Db.

340*t - 85 = 320 * t

340t - 320t = 85

20t = 85

t = 4.25 seconds

Now let's check our work:

rocket B is launched first and flies for a total of 4.25 seconds before exploding, at a speed of 320, that is:

320*4.25 = 1360

rocket A is launched 0.25 seconds after rocket B, so it's only flying for 4 seconds before exploding. at a speed of 340, that is:

340*4 = 1360

since both match, we know that 4.25 seconds after rocket B launches is when they will be at the same distance.