Question

Car A is moving twice as fast as car B. Car B starts 174. Meters in front of car A. If car A passes car B in 27 seconds, how fast is car B going.

Answer 1

Answer: The speed of car B is 6.44 m/s

Explanation:

Let's define:

Sa = Speed of car A.

Sb = Speed of car B.

I assume that at t = 0s, the position of car A is 0m (then at t = 0s, the position of car B is 174m)

Then we can write the positions of each car as:

Pa(t) = Sa*t

Pb(t) = Sb*t + 174m.

We also know that Sa = 2*Sb.

And we know that at t= 27s, car A passes car B.

Then at t = 27s, the positions of both cars must be the same:

Pa(27s) = Pb(27s)

Sa*27s = Sb*27s + 174m.

Now we can replace Sa by 2*Sb (from the above equation)

2*Sb*27s = Sb*27s + 174m

Now we can solve this for Sb.

2*Sb*27s - Sb*27s = 174m

Sb*27s = 174m

Sb = 174m/27s = 6.44m/s.

The speed of car B is 6.44 m/s