A dragster starts with zero velocity and completes a 404.5 m (0.2528 mile) run in 4.922 s. If the car had a constant acceleration, what would be its acceleration and final velocity?

33.40 m/s² and 164.38 m/s²

Explanation:

From the equation of motion,

s = ut + 1/2at² ... Equation 1.

Where s = distance, u = initial velocity, a = acceleration, t = time.

Given: s = 404.5 m, t = 4.922 s, u = 0 m/s.

Substitute into equation 1

404.5 = 0 (4.922) + 1/2 (4.922²a)

404.5 = 12.11a

a = 404.5/12.11

a = 33.40 m/s²

Also using

v² = u² + 2as ... Equation 2

Where v = final velocity.

Given: u = 0 m/s, a = 33.40 m/s². s = 404.5 m

Substitute into equation 2

v² = 0² + 2 (33.40) (404.5)

v² = 27020.6

v = √ (27020.6)

v = 164.38 m/s.

Hence acceleration and the final velocity is 33.40 m/s² and 164.38 m/s respectively

a. a=33.34ms⁻², V=164.4m/s

Explanation:

Since the dragster started with zero velocity, de determine the acceleration using of the equations of motion.

Below are the data given

Distance, s=404.5m,

time taken, t=4.922secs

Using the equation

S=ut+1/2at²

where u is the initial velocity and u=0

Making the acceleration the subject of the formula, we arrive at

a=2s/t²

a = (2*404.5) / 4.922²

a=33.34ms⁻².

To determine the velocity, we use

V=u+at

V=0+33.34ms⁻² * 4.922sec

V=164.4m/s