7. A formula 1 racecar is capable of very quick starts (with a launch control system, that prevents the wheels from spinning during rapid acceleration)). A 2011 Ferrari (driven by none other than Agent 001) is initially at rest, and undergoes a constant acceleration of 14 m/s^2 while in first gear. The car shifts to the second gear and has a constant acceleration of 13 m/s^2. While in second gear, the car travels 55m in 1.2 seconds before shifting to third gear. While in third gear, the acceleraton f the car is 9 m/s^2, and the car travels a distance of 35 m. Calculate
a) total distance
b) the total time
c) the final speed of the car
To calculate the total distance, total time, and final speed of the car, we need to break down the problem into three parts: first gear, second gear, and third gear.
a) To calculate the total distance, we need to find the distance traveled in each gear and then add them all together.
In the first gear:
Using the kinematic equation:
d = v_i * t + (1/2) * a * t^2
Since the car starts from rest, the initial velocity, v_i, is 0.
The time, t, is not given, so we need to find it using the equation:
a = (v_f - v_i) / t
Given:
a = 14 m/s^2
v_f = ?
v_i = 0
Rearranging the equation:
t = (v_f - v_i) / a
Substituting the values:
t = (v_f - 0) / 14
t = v_f / 14
Now, we can substitute this value of t into the first equation:
d = 0 * (v_f / 14) + (1/2) * 14 * (v_f / 14)^2
d = 0 + (1/2) * 14 * (v_f / 14)^2
d = (1/2) * (v_f / 14)^2
In the second gear:
Given:
a = 13 m/s^2
t = 1.2 seconds
d = 55 m
Using the kinematic equation:
d = v_i * t + (1/2) * a * t^2
Rearranging the equation:
v_i = (d - (1/2) * a * t^2) / t
Substituting the values:
v_i = (55 - (1/2) * 13 * (1.2^2)) / 1.2
v_i = 55 - (1/2) * 13 * (1.2)
v_i = 55 - (1/2) * 15.6
v_i = 55 - 7.8
v_i = 47.2
Now, let's calculate the distance traveled in second gear using the equation:
d = (v_i * t) + (1/2) * a * t^2
d = (47.2 * 1.2) + (1/2) * 13 * (1.2^2)
In the third gear:
Given:
a = 9 m/s^2
d = 35 m
Using the kinematic equation:
d = v_i * t + (1/2) * a * t^2
Let's calculate the initial velocity in the third gear using the equation:
v_i = sqrt(v_f^2 - 2 * a * d)
v_i = sqrt(v_f^2 - 2 * 9 * 35)
Now, we can substitute this value of v_i into the equation:
d = v_i * t + (1/2) * a * t^2
35 = v_i * t + (1/2) * 9 * t^2
b) To calculate the total time, we need to add the time taken in each gear.
In the first gear:
t_1 = v_f / 14 (calculated earlier)
In the second gear:
t_2 = 1.2 seconds (given)
In the third gear:
t_3 = t (to be calculated from the third gear equation)
Total time = t_1 + t_2 + t_3
c) To calculate the final speed of the car, we need to find the velocity at the end of each gear.
In the first gear:
v_f1 = v_f (as mentioned earlier)
In the second gear:
v_f2 = v_i (calculated earlier)
In the third gear:
v_f3 = v_f (as mentioned earlier)
Final speed = v_f (from the third gear equation)
Now, you can plug in the values and calculate the total distance, total time, and final speed of the car.