Question

Problem 8. A professional racecar driver buys a car that can accelerate at +5.9 m/s. The racer decides to race against another driver in a souped-up stock car. Both start from rest, but the stock-car driver leaves 1.0 s before the driver of the racecar. The stock car moves with a constant acceleration of +3.6 m/s. A. Find the time it takes the racecar driver to overtake the stock-car driver. B. Find the distance the two drivers travel before they are side by side. C. Find the velocities of both cars at the instant they are side by side.

Answer 1

A. After 3.57 seconds, the racecar will overtake the stock car

B. distance travelled = 37. 59 m

C. v₁ = 21.06 m/s

v₂ = 16.45 m/s

Step-by-step explanation:

Using, S = ut + 1/2at² where s is displacement, u is initial velocity, t is time, a is acceleration.

Car 1 has acceleration a = +5.9 m/s²; Car 2 has acceleration a = +3.6 m/s²

A. The time for the two cars to be directly side by side be t.

Beyond time, t, Car 1 will overtake Car 2.

Since both cars are starting from rest, initial velocity, u, = 0

S = 1/2at²

For car 1, time t = t - 1.0 since it started 1.0 second behind car 2

s₁ = 1/2 * 5.9 * (t - 1)²

s = 2.95t² - 5.9t + 2.95

For car 2; s₂ = 1/2 * 3.6 * t² = 1.8t²

at time t, the displacement of the two cars will be equal; s₁ = s₂

2.95t² - 5.9t + 2.95 = 1.8t²

1.15t² - 5.9t + 2.95 = 0

solving for t using the quadratic formula

t = (5.9 ± 4.61)/2.3

t = 4.57 s or 0.56 s

since t cannot be less than 1.0 s, t = 4.57 s

Time for car 1 = (4.57 - 1.0) s = 3.57 s

Therefore, after 3.57 seconds, the racecar will overtake the stock car

B. The distance travelled by both cars, s = 1/2 * 3.6 * 4.57² = 37.59 m OR

s = 1/2 * 5.9 * 3.57² = 37.59 m

distance travelled = 37. 59 m

C. Using v² = u² + 2as; since u = 0, v² = 2as

velocity of car 1; v₁² = 2 * 5.9 * 37.59

taking square root of both sides

v₁ = 21.06 m/s

velocity of car, v₂² = 2 * 3.6 * 37.59

taking square root of both sides

v₂ = 16.45 m/s