Brett stats from rest on his skateboard and accelerates at 0.5 m/s2 [N] for 12s, and then maintains his velocity for 15s before falling off his board. He slides along on his back until he comes to rest, accelerating at -2.4 m/s2.
a) Calculate the highest instantaneous velocity that Brett reaches.
b) How long did it take Brett to come to rest after falling?
c) Draw a velocity-time graph for the total time.
d) What is his displacement in the first 12s?
e) What is his displacement while maintaining a constant velocity?
f) What is his displacement after he fell?
g) What is Brett’s average velocity over the total time?
a) v = at
b) t = v/a
d) x = 1/2 at^2
e)x = vt
f) x = 1/2at^2
g) v = total dist/total time
a) To calculate the highest instantaneous velocity Brett reaches, we need to find the velocity at the end of the 12s acceleration period. The formula we can use is:
v = u + at
Where:
v = final velocity
u = initial velocity (in this case, 0 m/s since Brett starts from rest)
a = acceleration (0.5 m/s^2)
t = time (12s)
Plugging in the values, we get:
v = 0 + (0.5 m/s^2)(12s)
v = 6 m/s
Therefore, the highest instantaneous velocity Brett reaches is 6 m/s.
b) To calculate how long it took Brett to come to rest after falling, we need to find the time it takes for him to decelerate from his constant velocity to rest. The formula we can use is:
v = u + at
Where:
v = final velocity (0 m/s since he comes to rest)
u = initial velocity (in this case, the constant velocity during the 15s period)
a = acceleration (-2.4 m/s^2)
t = time (unknown)
Since he comes to rest, the final velocity is 0 m/s. Plugging in the values, we get:
0 = u + (-2.4 m/s^2)(t)
u = 2.4t
Since we don't know the initial velocity value, we can't calculate the time directly. However, we can use the fact that Brett maintains his velocity for 15s to find the initial velocity. Using the formula:
v = u + at
Where:
v = final velocity (the constant velocity during the 15s period)
u = initial velocity (unknown)
a = acceleration (0 m/s^2 since he maintains the velocity)
t = time (15s)
Plugging in the values, we get:
v = u + (0 m/s^2)(15s)
v = u
Therefore, the initial velocity is equal to the constant velocity during the 15s period.
Now we have two equations:
1) 0 = u - 2.4t
2) v = u
By substituting equation 2 into equation 1, we get:
0 = v - 2.4t
Now, we can solve for t:
t = v / 2.4
Plugging in the known values for v (the constant velocity during the 15s period), we can calculate t.
c) To draw a velocity-time graph for the total time, we need to consider the different periods of motion.
From 0s to 12s, Brett is accelerating at 0.5 m/s^2, so the velocity-time graph would be a straight line with a positive slope (representing the increasing velocity over time).
From 12s to 27s, Brett maintains a constant velocity, so the velocity-time graph would be a straight horizontal line (representing a constant velocity).
From 27s until Brett comes to rest, he decelerates at -2.4 m/s^2, so the velocity-time graph would be a straight line with a negative slope (representing the decreasing velocity over time).
d) To calculate Brett's displacement in the first 12s, we can use the formula:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity (0 m/s since Brett starts from rest)
t = time (12s)
a = acceleration (0.5 m/s^2 during the first 12s)
Plugging in the values, we get:
s = 0(12s) + (1/2)(0.5 m/s^2)(12s)^2
s = 0 + (1/2)(0.5 m/s^2)(144s^2)
s = (1/2)(0.5 m/s^2)(144s^2)
s = 0.5(0.5 m/s^2)(144s^2)
s = 0.25(144s^2)
s = 36s^2
Therefore, Brett's displacement in the first 12s is 36s^2.
e) While maintaining a constant velocity, Brett's displacement can be calculated using the formula:
s = vt
Where:
s = displacement
v = velocity (constant velocity during the 15s period)
t = time (15s)
Plugging in the values, we get:
s = v(15s)
Therefore, Brett's displacement while maintaining a constant velocity is v(15s).
f) Brett's displacement after he fell can be calculated using the formula:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity (the velocity after he fell)
t = time (unknown)
a = acceleration (-2.4 m/s^2 during the deceleration period)
Since Brett's final velocity is 0 m/s, we can use the formula:
v = u + at
Where:
v = final velocity (0 m/s)
u = initial velocity (the velocity after he fell)
a = acceleration (-2.4 m/s^2 during the deceleration period)
t = time (unknown)
Plugging in the values, we get:
0 = u + (-2.4 m/s^2)(t)
u = 2.4t
Now, we can use this equation in the displacement formula:
s = (2.4t)(t) + (1/2)(-2.4 m/s^2)(t)^2
s = 2.4t^2 - 1.2t^2
s = 1.2t^2
Therefore, Brett's displacement after he fell is 1.2t^2.
g) To calculate Brett's average velocity over the total time, we need to divide the total displacement by the total time.
Total displacement = displacement in the first 12s + displacement while maintaining constant velocity + displacement after he fell
Total time = time during the first 12s + time while maintaining constant velocity + time after he fell
Using the previously calculated formulas, find the values of the displacements and times. Then, divide the total displacement by the total time to get Brett's average velocity over the total time.