A drag racer, starting from rest, travels 6.0 m in 1.0 s. Suppose the car contin

Question

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.

Explanation / Answer

Here m

let the accelearation of the car is a

for the first 1 s

Using second equation of motion

d= 0.5 * a * t^2

6 = 0.5 * a * 1^2

a = 12 m/s^2

Now , for the distance travelled in total (4 + 1) = 5 s

distance from the starting point = 0.5 * a * t^2

distance from the starting point = 0.5 * 12 * 5^2

distance from the starting point = 150 m

hence , the distance from the starting point of car is 150 m

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