A car starts from rest and accelerates with a constant acceleration of 1.00 m/s2 for 3.00 s. The car continues for 5.00 s at constant velocity. How far has the car traveled from its starting point?

19.5 m

Explanation:

Using the equation of motion,

For the first 3 seconds,

s = ut +1/2at² Equation 1

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

Note: Since the car starts from rest, u = 0 m/s.

Given: a = 1.0 m/s, t = 3 s, u = 0 m/s

Substitute into equation 1

s = 0×3+1/2(3)²(1)

s = 9/2

s = 4.5

In the remaining five seconds, at a constant velocity,

s' = vt Equation 2

Where v = velocity, t = time.

Recall,

v = u+at

v = 0+1(3)

v = 3 m/s.

Also, t = 5 s.

Substitute into equation 2

s' = 3(5)

s' = 15 m.

Total distance = 15+4.5

Total distance = 19.5 m.

Hence the car travels 19.5 m

prolly a beacuse the momentuem the car has increases while going down hill