A young man named Peter buys a sports car that can accelerate at the rate of 4.90 m/s2. He decides to test the car by racing with another speedster, John. Both start from rest, but experienced John leaves the starting line 1.00 s before Peter. If John moves with a constant acceleration of 3.50 m/s2, and Peter maintains an acceleration of 4.90 m/s2, find (a) the time it takes Peter to overtake John, (b) the distance Peter travels before he catches John, and (c) the speeds of both cars at the instant Peter overtakes John.

a) 6.46 s

b) 73.0 m

c) 22.6 m/s, 26.7 m/s

Explanation:

The position can be found with constant acceleration equation:

x = x₀ + v₀ t + ½ at²

where x is the final position,

x₀ is the initial position,

v₀ is the initial velocity,

a is the acceleration,

and t is the time.

For John:

x₀ = 0 m

v₀ = 0 m/s

a = 3.50 m/s²

For Peter:

x₀ = 0 m

v₀ = 0 m/s

a = 4.90 m/s²

John's position at time t is:

x = 0 + (0) t + ½ (3.50) t²

x = 1.75 t²

Peter starts 1 second after John. Peter's position at time t-1 is:

x = 0 + (0) (t-1) + ½ (4.90) (t-1) ²

x = 2.45 (t-1) ²

When Peter overtakes John, they have the same position:

1.75 t² = 2.45 (t-1) ²

1.75 t² = 2.45 (t² - 2t + 1)

1.75 t² = 2.45 t² - 4.90 t + 2.45

0 = 0.70 t² - 4.90 t + 2.45

0 = 2 t² - 14 t + 7

t = [ 14 ± √ (169 - 4 (2) (7)) ] / 4

t = 0.54, 6.46

Since t > 1, Peter overtakes John 6.46 seconds after the race starts, which means he drives for 5.46 seconds.

The distance Peter travels is:

x = 2.45 (6.46 - 1) ²

x = 73.0

Peter travels 73.0 meters.

The speed that John reaches is:

v = (3.50 m/s²) (6.46 s)

v = 22.6 m/s

The speed that Peter reaches is:

v = (4.90 m/s²) (5.46 s)

v = 26.7 m/s