A jogger accelerates from rest to 5.78 m/s in 3.51 s. A car accelerates from 19.7 to 32.1 m/s also in 3.51 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 3.51 s?
To solve this problem, we need to first find the acceleration of the jogger and the car using the given information. Then, we can calculate the distance traveled by each using the acceleration and time.
(a) To find the magnitude of the acceleration of the jogger, we can use the equation:
a = (v - u) / t
where:
a = acceleration
v = final velocity
u = initial velocity
t = time taken
Given:
u = 0 (rest)
v = 5.78 m/s
t = 3.51 s
Substituting the values into the equation, we get:
a = (5.78 - 0) / 3.51
a = 1.645 m/s^2
Therefore, the magnitude of the acceleration of the jogger is 1.645 m/s^2.
(b) To find the magnitude of the acceleration of the car, we use the same equation as before:
a = (v - u) / t
Given:
u = 19.7 m/s
v = 32.1 m/s
t = 3.51 s
Substituting the values into the equation, we get:
a = (32.1 - 19.7) / 3.51
a = 3.527 m/s^2
Therefore, the magnitude of the acceleration of the car is 3.527 m/s^2.
(c) To find how much further the car travels than the jogger during the 3.51 s, we can use the equations of motion. The distance covered by an object undergoing constant acceleration can be calculated using the equation:
s = ut + (1/2)at^2
where:
s = distance
u = initial velocity
t = time taken
a = acceleration
For the jogger:
u = 0
t = 3.51 s
a = 1.645 m/s^2
Substituting the values into the equation, we get:
s_jogger = (0)(3.51) + (1/2)(1.645)(3.51)^2
s_jogger = 8.649 m
For the car:
u = 19.7 m/s
t = 3.51 s
a = 3.527 m/s^2
Substituting the values into the equation, we get:
s_car = (19.7)(3.51) + (1/2)(3.527)(3.51)^2
s_car = 138.43 m
Therefore, the car travels 138.43 m - 8.649 m = 129.781 m further than the jogger during the 3.51 s.