physics help?

The steps are to determine the car’s maximum velocity and the distance it moved during the 7.4 seconds.

Eq #1,vf = vi + a * t, vi = 0
vf = 2.2 * 7.4 = 16.28 m/s

Eq#2, d = ½ * (vi + vf) * t, vi = 0
d = ½ * 16.28 * 7.4 = 60.236 meters

Let’s use the same equations for the braking time. This time the initial velocity is 16.28 m/s.

vf = 16.28 + -1.8 * 2.4 = 11.96 m/s
d = ½ * (16.28 + 11.96) * 2.4 = 33.888 meters

OR

d = vi * t + ½ * a * t^2
d = 16.28 * 2.4 + ½ * -1.8 * 2.4^2 = 33.888 meters
This is exactly the same answer.


The total distance is the sum of these two distances.

d = 60.236 + 33.888 = 94.124 meters
I hope this is helpful for you.

Consider the journey of acceleration :
Initial velocity, u = 0 m/s
Time taken, t = 7.4 s
Acceleration, a = 2.2 m/s²

v = u + at
Final velocity when accelerating, v = 0 + 2.2 * 7.4 = 2.2 * 7.4 m/s

s = ut + (1/2)at²
Distance traveled when accelerating, s = 0 + (1/2) * 2.2 * 7.4² = 1.1 * 7.4² m

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Consider the journey of deceleration :
Initial velocity, u = 2.2 * 7.4 m/s
Final velocity, v = 0 m/s
Time taken, t = 2.4 s

s = [(u + v)/2] t
Distance traveled when decelerating, s = [(2.2 * 7.4 + 0)/2] * 2.4 = 1.1 * 7.4 * 2.4 m

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Distance from its start = (1.1 * 7.4² + 1.1 * 7.4 * 2.4) m = 80 m

At the end of the braking period the car will be going at (7.4 x 2.2) - (1.8 x 2.4)

i got a 67% on my physics test :(