Question

Starting from rest, a dragster travels a straight 1/4 mi racetrack in 6.70 s with constant acceleration. What is its velocity when it crosses the finish line?

Answer 1

Its velocity when it crosses the finish line is 119.40 m/s

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = ?

        Time, t = 6.70 s      

        Displacement, s = 1/4 mi = 0.25 mi = 400 m

     Substituting

                      s = ut + 0.5 at²

                      400 = 0 x 6.70 + 0.5 x a x 6.70²

                      a = 17.82 m/s²

     Now we have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = ?

     Time, t = 6.7 s

     Acceleration, a = 17.82 m/s²

     Substituting

                      v = u + at  

                      v = 0 + 17.82 x 6.7

                      v = 119.40 m/s

Its velocity when it crosses the finish line is 119.40 m/s

Answer 2

The final velocity of the dragster as it crosses the finish line is 120.1m/s

Given the data in the question;

Since the dragster was initially at rest;

• Initial velocity; $u = 0$

• Distance travelled; $s = \frac{1}{4}miles = 0.25miles = 402.336m$

• Time taken; $t = 6.70s$

Final velocity; $v = \ ?$

First we determine the acceleration of the dragster, using the second equation of motion:

$s = ut + \frac{1}{2}at^2$

We substitute our values into the equation

$402.336m = [0 * 6.70s] + [ \frac{1}{2} * a * (6.70s)^2 ]\\\\402.336m = \frac{1}{2} * a * 44.89s^2 \\\\402.336m = 22.445s^2 * a\\\\a = \frac{402.336m}{22.445s^2}\\\\a = 17.925m/s^2$

Now, to determine the final velocity of the dragster as it crosses the finish line, we use the first equation of motion:

$v = u + at$

We substitute our values into the equation

$v = 0 + [ 17.925m/s^2\ *\ 6.70s ]\\\\v = 17.925m/s^2\ *\ 6.70s\\\\v = 120.1m/s$

Therefore, the final velocity of the dragster as it crosses the finish line is 120.1m/s.

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