Answer:
a = 0.9955 m/s²
t = 15 s
Explanation:
A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 15.0 m/s when it reaches the end of the ramp, which has length 113 m.
What is the acceleration of the car?
How much time does it take the car to travel the length of the ramp?
Given
v₀ = 0 m/s
v = 15 m/s
x = 113 m
a = ?
t = ?
In order to get the acceleration, we can apply the formula
v² = v₀² + 2ax
⇒ a = (v²-v₀²)/(2x) = ((15 m/s)²-(0 m/s)²)/(2*113 m)
⇒ a = 0.9955 m/s²
Then we use the following equation in order to get t:
v = v₀ + at
⇒ t = (v - v₀)/a = ((15 m/s) - (0 m/s))/(0.9955 m/s²)
⇒ t = 15 s
