a motorcycle initially moving at +32 m/s accelerates at -4.0 m/s for 3.0s. a car initially at rest accelerates at 4.0 m/s for 5.25 s.
a) Which vehicle will be going faster at the end of each given time, and how much faster?
b) how far does each vehicle travel during their respective times?
32 m/sec will be decreased at for 3.0 sec.
32-12 = 20 m/s
4.0 m/sec times 5.25 sec. = 21 m/sec
1m/sec faster.
b) d= rt
You could use the average rate by computing the beginning and ending speeds
To determine which vehicle will be going faster at the end of each time period and by how much, we need to calculate the final velocities of both the motorcycle and the car. We can use the equations of motion to find the final velocities.
The equations of motion we'll be using are:
(1) v = u + at - for calculating final velocity
(2) s = ut + (1/2)at^2 - for calculating distance traveled
a) Calculating final velocities:
For the motorcycle:
Initial velocity (u1) = +32 m/s (positive because it's moving in the positive direction)
Acceleration (a1) = -4.0 m/s^2 (negative because it's decelerating)
Time (t1) = 3.0 s
Using equation (1) for the motorcycle:
v1 = u1 + a1 * t1
v1 = 32 m/s + (-4.0 m/s^2) * 3.0 s
v1 = 32 m/s - 12 m/s
v1 = 20 m/s (positive)
For the car:
Initial velocity (u2) = 0 m/s (as it's at rest)
Acceleration (a2) = 4.0 m/s^2
Time (t2) = 5.25 s
Using equation (1) for the car:
v2 = u2 + a2 * t2
v2 = 0 m/s + (4.0 m/s^2) * 5.25 s
v2 = 0 m/s + 21 m/s
v2 = 21 m/s
Therefore, at the end of the given time periods:
- The motorcycle will be going faster with a velocity of 20 m/s.
- The car will be going faster with a velocity of 21 m/s.
The car will be faster by 21 - 20 = 1 m/s.
b) Calculating distances traveled:
To find the distances traveled by each vehicle during their respective times, we'll use equation (2).
For the motorcycle:
Distance (s1) = initial velocity (u1) * time (t1) + (1/2) * acceleration (a1) * time^2 (t1)
s1 = (32 m/s) * (3.0 s) + (1/2) * (-4.0 m/s^2) * (3.0 s)^2
s1 = 96 m - 18 m
s1 = 78 m
For the car:
Distance (s2) = initial velocity (u2) * time (t2) + (1/2) * acceleration (a2) * time^2 (t2)
s2 = (0 m/s) * (5.25 s) + (1/2) * (4.0 m/s^2) * (5.25 s)^2
s2 = 0 m + 44.1 m
s2 = 44.1 m
Therefore, the motorcycle travels a distance of 78 m, and the car travels a distance of 44.1 m during their respective time periods.