A jogger accelerates from rest to 2.86 m/s in 3.38 s. A car accelerates from 22.9 to 35.0 m/s also in 3.38 s. (a) Find the magnitude of the acceleration of the jogger. (b) Determine the magnitude of the acceleration of the car. (c) How much further does the car travel than the jogger during the 3.38 s?
a = ∆v/t
the distance for each is s = vt + 1/2 at^2
a = change in v / change in time
a) (2.86 - 0) / 3.38 = 0.846 m/s^2 (pitiful, g = about 10)
b) (35.0 - 22.9) / 3.38 = 3.58 m/s^2
c) Jogger goes (1/2)a t^2 = (1/2)(0.846)(11.4) = 4.83 meters
car goes vt+(1/2)at^2 =22.9*3.38+(1/2)(3.58)(11.4) = 77.4+20.4= 97.8 meters
difference = 97.8 - 4.8 = 93 meters
To find the magnitude of acceleration, we can use the formula:
acceleration = (final velocity - initial velocity) / time
(a) For the jogger:
The initial velocity of the jogger is 0 m/s (since the jogger starts from rest) and the final velocity is 2.86 m/s. The time taken is 3.38 s. Plugging these values into the formula, we get:
acceleration of the jogger = (2.86 m/s - 0 m/s) / 3.38 s
acceleration of the jogger = 2.86 m/s / 3.38 s
acceleration of the jogger = 0.847 m/s^2
Therefore, the magnitude of the acceleration of the jogger is 0.847 m/s^2.
(b) For the car:
The initial velocity of the car is 22.9 m/s and the final velocity is 35.0 m/s. The time taken is 3.38 s. Plugging these values into the formula, we get:
acceleration of the car = (35.0 m/s - 22.9 m/s) / 3.38 s
acceleration of the car = 12.1 m/s / 3.38 s
acceleration of the car = 3.58 m/s^2
Therefore, the magnitude of the acceleration of the car is 3.58 m/s^2.
(c) To find out how much further the car travels than the jogger during the 3.38 s, we need to calculate the distances covered by each of them. The formula for distance covered with constant acceleration is:
distance = initial velocity * time + (1/2) * acceleration * time^2
For the jogger:
Using the initial velocity as 0 m/s and the acceleration as 0.847 m/s^2, we can calculate the distance covered by the jogger as:
distance = 0 * 3.38 s + (1/2) * 0.847 m/s^2 * (3.38 s)^2
distance = 0 + (1/2) * 0.847 m/s^2 * 11.41 s^2
distance = (1/2) * 9.665 s^2
distance = 4.833 m
For the car:
Using the initial velocity as 22.9 m/s and the acceleration as 3.58 m/s^2, we can calculate the distance covered by the car as:
distance = 22.9 m/s * 3.38 s + (1/2) * 3.58 m/s^2 * (3.38 s)^2
distance = 77.462 m + (1/2) * 3.58 m/s^2 * 11.41 s^2
distance = 77.462 m + (1/2) * 40.961 m
distance = 77.462 m + 20.4805 m
distance = 97.9425 m
Therefore, the car travels 97.9425 m - 4.833 m = 93.1095 m further than the jogger during the 3.38 s.