A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t)= α t2− β t3, where α = 1.43m/s2 and β = 4.60×10−2m/s3 .
1)Calculate the average velocity of the car for the time interval t=0 to t1 = 2.04s
2)Calculate the average velocity of the car for the time interval t=0 to t2 = 4.09s .
3)Calculate the average velocity of the car for the time interval t1 = 2.04s to t2 = 4.09s .
velocity=d distance/dtime= d/dt (a t^2-Bt^3)
velocity=2A*t -3Bt^2
a) average= (v(2.04)-v(o))/2.04 where
v(2.04)= 2A*2.04-3B*2.04^2 where A, B are given.
A small rock is thrown vertically upward with a speed of 12.0 m/s from the edge of the roof of a 23.0 m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.
Part A
What is the speed of the rock just before it hits the street?
Express your answer with the appropriate units.
To calculate the average velocity of the car for different time intervals, we need to find the displacement during that time interval and divide it by the duration of the interval.
1) Average velocity for the time interval t = 0 to t1 = 2.04s:
To find the displacement during this time interval, we subtract the initial position from the final position. The initial position is x(t = 0) and the final position is x(t1 = 2.04s).
Let's calculate it:
x(t = 0) = αt^2 - βt^3
= (1.43 m/s^2)(0^2) - (4.60×10^-2 m/s^3)(0^3)
= 0
x(t1 = 2.04s) = αt^2 - βt^3
= (1.43 m/s^2)(2.04^2) - (4.60×10^-2 m/s^3)(2.04^3)
= 5.834 m
The displacement during this time interval is the difference between the two positions:
Δx = x(t1 = 2.04s) - x(t = 0)
= 5.834 m - 0
= 5.834 m
The duration of the interval is t1 - t = 2.04s - 0s
= 2.04s
Therefore, the average velocity for the time interval t = 0 to t1 = 2.04s is:
Average velocity = Δx / Δt
= 5.834 m / 2.04s
= 2.859 m/s
So, the average velocity of the car for the time interval t = 0 to t1 = 2.04s is 2.859 m/s.
2) Average velocity for the time interval t = 0 to t2 = 4.09s:
Using the same approach as above, we need to calculate the displacement during this time interval and divide it by the duration of the interval.
x(t2 = 4.09s) = αt^2 - βt^3
= (1.43 m/s^2)(4.09^2) - (4.60×10^-2 m/s^3)(4.09^3)
= 30.767 m
The displacement during this time interval is the difference between the two positions:
Δx = x(t2 = 4.09s) - x(t = 0)
= 30.767 m - 0
= 30.767 m
The duration of the interval is t2 - t = 4.09s - 0s
= 4.09s
Therefore, the average velocity for the time interval t = 0 to t2 = 4.09s is:
Average velocity = Δx / Δt
= 30.767 m / 4.09s
= 7.518 m/s
So, the average velocity of the car for the time interval t = 0 to t2 = 4.09s is 7.518 m/s.
3) Average velocity for the time interval t1 = 2.04s to t2 = 4.09s:
Now, we need to calculate the displacement during this time interval and divide it by the duration of the interval.
Using the same formula, plug in the values for t2 and t1:
x(t2 = 4.09s) = αt^2 - βt^3
= (1.43 m/s^2)(4.09^2) - (4.60×10^-2 m/s^3)(4.09^3)
= 30.767 m
x(t1 = 2.04s) = αt^2 - βt^3
= (1.43 m/s^2)(2.04^2) - (4.60×10^-2 m/s^3)(2.04^3)
= 5.834 m
The displacement during this time interval is the difference between the two positions:
Δx = x(t2 = 4.09s) - x(t1 = 2.04s)
= 30.767 m - 5.834 m
= 24.933 m
The duration of the interval is t2 - t1 = 4.09s - 2.04s
= 2.05s
Therefore, the average velocity for the time interval t1 = 2.04s to t2 = 4.09s is:
Average velocity = Δx / Δt
= 24.933 m / 2.05s
= 12.142 m/s
So, the average velocity of the car for the time interval t1 = 2.04s to t2 = 4.09s is 12.142 m/s.
To calculate the average velocity of the car for a given time interval, we need to use the formula:
Average Velocity = (change in distance) / (change in time)
1) For the time interval t = 0 to t1 = 2.04s:
- Calculate the change in distance: Δx = x(t1) - x(0)
- Substitute the values into the equation x(t) = αt^2 - βt^3:
Δx = αt1^2 - βt1^3 - αt0^2 + βt0^3
- Calculate the change in time: Δt = t1 - t0 = t1 - 0
- Substitute the given values of α, β, t1, and t0:
Δt = t1 - t0 = 2.04s - 0s
- Divide the change in distance by the change in time to find the average velocity:
Average Velocity = Δx / Δt
2) For the time interval t = 0 to t2 = 4.09s, follow the same steps as above.
3) For the time interval t1 = 2.04s to t2 = 4.09s:
- Calculate the change in distance: Δx = x(t2) - x(t1)
- Calculate the change in time: Δt = t2 - t1
- Divide the change in distance by the change in time to find the average velocity.
By following these steps, you can calculate the average velocity of the car for each given time interval.