Physics word problems

Question

A race driver has made a pit stop to refuel. After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is 5.5 m/s?; after 4.2 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 73.0 m/s overtakes and passes the entering car. The entering car maintains its acceleration. How much time is required for the entering car to catch up with the other car?

Answer 1

Consider the following diagram representing the problem:

Car 1--->------------------>

Car 2--->------------------>

The first arrowhead to the right of car 1 represents the car entering the speedway after its initial acceleration. The second arrowhead to the right of car 1 represents the point on the speedway when it has caught up with car 2.

The first arrowhead to the right of car 1 is the point when it is in the same position as car 1 when it first enters the speedway. The second arrowhead is car 2s position on the speedway where car 1 catches up with it.

Note that the distance between the first and second arrowheads for each car is the same. Let that distance be S.

We need to find the time for either car to travel that distance.

S1 = V1₀ t + ½ a1 t² where S1 is the distance travelled by car 1, V1₀ is it velocity after accelerating to the speedway, and a1 is its acceleration.

S2 = V2₀ t + ½ a t² where V2₀ is the constant velocity of car 2. As car 2 is not accelerating, it’s a = 0, so S2 = V2₀ t.

S1 = S2, so V1₀ t + ½ a1 t² = V2₀ t.

The unknowns here are V1₀ and t. We can calculate V1₀ since V1₀ = 5.5 m/s² and time is given as 4.2 s. So, V1₀ = 5.5 m/s² x 4.2 s = 23.1 m/s.

Substituting this into the equation above, we get 23.1 m/s x t + ½ x 5.5 m/s²x t² = 73.0 m/s x t. 

 t is the only unknown, so It can be solved for. A little algebra gives us

2.75 t² -49.9 t = 0 and t (2.75t – 49.9) = 0 after factoring out t. Set the expression in parentheses = 0 to get t = 18.1 s.

There is a lot of math here, so check it yourself.

Answer 2

Let's first figure out how fast car #1 was going as it entered the track:

v_f = v_i + at where v_f = final velocity (the speed the car was going as it entered the track), v_i = initial velocity (which was zero because the car started from a stop), "a" = acceleration, and "t" = time.

v_f = 0 + (5.5)(4.2) = 23.1 m/s

We can now use the location on the track where car #2 passed car #1 as our starting position. You want to know the distance from this starting position that car #1 travels and the distance from that position that car #2 travels and you want those distances to be the same (the location where car #1 catches up).

For car #1, the distance traveled can be calculated from:

x = v_it + (1/2)at² where v_i = 23.1, "t" is the time it takes to catch up, and "a" is the acceleration:

x = 23.1t + (1/2)(5.5)t²

x = 23.1t + 2.75t²

For car #2:

x = vt = (73.0)t

Since the two distances are the same:

23.1t + 2.75t² = 73t

2.75t² - 49.9t = 0

t(2.75t - 49.9) = 0

t = 0 and t = 18.1 seconds

When t = 0, the 2 cars where at the same location (this was the place car #2 passed car #1) and when t = 18.1 s, the cars again where at the same location (this is where car #1 caught up to car #2)