1. A woman driving at the 45 mi/hour speed limit on the entrance ramp to the highway accelerates at a constant rate and reaches the highway speed limit of 65 mi/hour in 6.00 s. What distance does the car travel during that acceleration? (Make the simplifying assumption that she is traveling in a straight line and be careful with your units)

Respuesta :

Answer:

s = 147.54 m

Explanation:

given,

initial velocity,u = 45 mi/h

1 mph = 0.44704 m/s

45 mph = 45 x 0.44704 = 20.12 m/s

final velocity, v = 65 mi/h

            v = 65 x 0.44704 = 29.06 m/s

time, t = 6 s

acceleration, [tex]a = \dfrac{v-u}{t}[/tex]

[tex]a = \dfrac{29.06-20.12}{6}[/tex]

 a = 1.49 m/s²

distance travel by the car

using equation of motion

v² = u² + 2 a s

29.06² = 20.12² + 2 x 1.49 x s

2.98 s = 439.6692

s = 147.54 m

distance traveled by the car is equal to 147.54 m