Question

A 873-kg (1930-lb) dragster, starting from rest completes a 401.4-m (0.2509-mile) run in 4.945 s. If the car had a constant acceleration, what would be its acceleration and final velocity

Answer 1

To solve this problem it is necessary to apply the kinematic equations of motion.

By definition we know that the position of a body is given by

$x=x_0+v_0t+at^2$

Where

$x_0 =$ Initial position

$v_0 =$ Initial velocity

a = Acceleration

t= time

And the velocity can be expressed as,

$v_f = v_0 + at$

Where,

$v_f = Final velocity$

For our case we have that there is neither initial position nor initial velocity, then

$x= at^2$

With our values we have $x = 401.4m, t=4.945s$, rearranging to find a,

$a=\frac{x}{t^2}$

$a = \frac{ 401.4}{4.945^2}$

$a = 16.41m/s^2$

Therefore the final velocity would be

$v_f = v_0 + at$

$v_f = 0 + (16.41)(4.945)$

$v_f = 81.14m/s$

Therefore the final velocity is 81.14m/s