A drag racer, starting from rest, travels 6.0 m in 1.0 s. Suppose the car continues this acceleration for an additional 4.0 s. How far from the starting line will the car be? We assume that the acceleration is constant, and the initial speed is zero, so the displacement will scale as the square of the time. After 1.0 s, the car has traveled 6.0 m; after another 4.0 s, a total of 5.0 s will have elapsed. The initial elapsed time was 1.0 s, so the elapsed time increases by a factor of 5. The displacement thus increases by a factor of 5^2, or 25. The total displacement is Delta x = 25 (6.0 m) = 150 m This is a big distance in a short time, but drag racing is a fast sport, so our answer makes sense.

Question

Physics

Alejandro Cisneros

7 September, 06:12

A drag racer, starting from rest, travels 6.0 m in 1.0 s. Suppose the car continues this acceleration for an additional 4.0 s. How far from the starting line will the car be? We assume that the acceleration is constant, and the initial speed is zero, so the displacement will scale as the square of the time. After 1.0 s, the car has traveled 6.0 m; after another 4.0 s, a total of 5.0 s will have elapsed. The initial elapsed time was 1.0 s, so the elapsed time increases by a factor of 5. The displacement thus increases by a factor of 5^2, or 25. The total displacement is Delta x = 25 (6.0 m) = 150 m This is a big distance in a short time, but drag racing is a fast sport, so our answer makes sense.

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Answers (1)

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Rylan Skinner · Answer 1

x = 150.0 m

Explanation:

If the acceleration is constant, we can find the value of the acceleration, starting from rest, applying the following kinematic equation:

x = 1/2*a*t² = 6.0 m

Solving for a:

a = 2*x / t² = 2*6.0 m / 1.0s² = 12 m/s²

Now, during the following 4.0 s, the car continues moving with this acceleration, but its initial velocity is not zero anymore, but the speed at 1.0 s, which is just 12 m/s (as it accelerates 12 m/s each second), so we can write again the same kinematic equation, taking into account that initial velocity for the second part, as follows:

x = x₀ + v₀*t + 1/2*a*t² = 6.0 m + 12m/s*4.0s + 1/2*12 m/s² * (4.0) s² = 150.0 m

⇒ x = 150.0 m