A race car is able to accelerate at a rate of 14m/s2. After exiting a turn at a speed of 15m/s, the car enters a straightaway and accelerates to a speed of 96m/s just as it reaches the end of the straightaway and enters the next turn.
A)how long did it take the car to accelerate to 96m/s.
b) how long is the straightaway, from the point where the car began accelerating the to point where it entered the next turn?
a. Vf = Vo + at,
t = (Vf-Vo)/a = (96-15) / 14 = 5.79s.
b. Vf^2 = Vo^2 + 2ad,
d = (Vf^2-Vo^2)/2a,
d = (9216-225) / 28 = 321m.
To answer both parts of the question, we can use the equations of motion.
a) To find the time it took the car to accelerate to 96m/s, we can use the equation:
v = u + at
Where:
v = final velocity (96m/s)
u = initial velocity (15m/s)
a = acceleration (14m/s^2)
t = time
Rearranging the equation to solve for time (t), we have:
t = (v - u) / a
Substituting the given values:
t = (96 - 15) / 14
t = 81 / 14
t ≈ 5.79 seconds
So, it took approximately 5.79 seconds for the car to accelerate to 96m/s.
b) To find the length of the straightaway, we can use the equation:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity (15m/s)
t = time (5.79 seconds)
a = acceleration (14m/s^2)
Rearranging the equation to solve for distance (s), we have:
s = ut + (1/2)at^2
Substituting the given values:
s = (15 * 5.79) + (1/2) * 14 * (5.79^2)
s = 86.85 + 417.5226
s ≈ 504.37 meters
So, the length of the straightaway, from the point where the car began accelerating to the point where it entered the next turn, is approximately 504.37 meters.