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 = 1.40m/s2 and β = 5.05×10−2m/s3
A) What is the average velocity of the car for the time interval t=0 to t1=1.91s?
B) What is the average velocity of the car for the time interval t=0 to t2 = 3.99s?
C) Calculate the average velocity of the car for the time interval t1 = 1.91s to t2 = 3.99s
I don't know where to even start.
x = 1.40 t^2 - .0505 t^3
at t = 0, x = 0
at t = 1.91, x =1.40 * 1.91^2 - .0505 * 1.91^3 = 4.755 meters
so
4.755 meters in 1.91 seconds = 2.5 m/s
do the others the same way, find out how far it went in the time interval and divide by the time interval
Can you help me get started on C? I'm stuck there.
Well, looks like you're on the road to confusion. Luckily, Clowns Bot is here to help guide you!
To find the average velocity, we need to calculate the change in position divided by the change in time. In this case, we're given the position function x(t) for the car, which is x(t) = αt^2 - βt^3.
To find the change in position, we simply subtract the initial position from the final position. So let's dive in and calculate the average velocities!
A) To find the average velocity from t = 0 to t1 = 1.91s, we'll need to calculate the change in position and divide it by the change in time. Let's call the initial position x0 (x at t=0) and the final position x1 (x at t1).
x0 = α*(0)^2 - β*(0)^3 = 0
x1 = α*(1.91)^2 - β*(1.91)^3
The change in position is: Δx = x1 - x0
Now, let's calculate the average velocity by dividing the change in position by the change in time: average velocity = (Δx) / (t1 - t0)
B) Similar to part A, we'll calculate the change in position and divide it by the change in time. The initial position x0 is still 0, and the final position x2 will be:
x0 = α*(0)^2 - β*(0)^3 = 0
x2 = α*(3.99)^2 - β*(3.99)^3
The change in position is: Δx = x2 - x0
Now, calculate the average velocity: average velocity = (Δx) / (t2 - t0)
C) Finally, to find the average velocity from t1 = 1.91s to t2 = 3.99s, we'll again calculate the change in position and divide it by the change in time.
The initial position x1 is equal to the final position x1 calculated earlier, and the final position x2 is still calculated as:
x1 = α*(1.91)^2 - β*(1.91)^3
x2 = α*(3.99)^2 - β*(3.99)^3
The change in position is: Δx = x2 - x1
Now, calculate the average velocity: average velocity = (Δx) / (t2 - t1)
Remember to plug in the values of α and β given in the problem.
I hope my explanation didn't drive you too crazy! Keep calm and calculate on!
To find the average velocity, we need to calculate the change in position divided by the change in time. Let's break down the steps to find the average velocity for each part.
A) Average velocity for the time interval t = 0 to t1 = 1.91s.
1. Substitute t = 0 into the equation x(t) = αt^2 - βt^3 to find the initial position x0.
x(0) = α(0)^2 - β(0)^3 = 0
So, the initial position x0 = 0.
2. Substitute t = t1 into the equation to find the final position x1.
x(t1) = α(t1)^2 - β(t1)^3
3. Calculate the change in position Δx = x1 - x0.
4. Calculate the change in time Δt = t1 - 0.
5. Finally, divide Δx by Δt to find the average velocity.
B) Average velocity for the time interval t = 0 to t2 = 3.99s.
1. Substitute t = t2 into the equation x(t) = αt^2 - βt^3 to find the final position x2.
x(t2) = α(t2)^2 - β(t2)^3
2. Calculate the change in position Δx = x2 - x0. (Since we started at t = 0, x0 = 0)
3. Calculate the change in time Δt = t2 - 0.
4. Divide Δx by Δt to find the average velocity.
C) Average velocity for the time interval t1 = 1.91s to t2 = 3.99s.
1. Substitute t = t1 into the equation x(t) = αt^2 - βt^3 to find the position at t1.
x(t1) = α(t1)^2 - β(t1)^3
2. Substitute t = t2 into the equation to find the position at t2.
x(t2) = α(t2)^2 - β(t2)^3
3. Calculate the change in position Δx = x(t2) - x(t1).
4. Calculate the change in time Δt = t2 - t1.
5. Divide Δx by Δt to find the average velocity.
Remember to substitute the given values of α and β into the equations. Let's solve each of these steps for the specific average velocities.
Two insulating spheres have radii 0.300 cm and 0.500 cm, masses 0.500 kg and 0.700 kg, and uniformly distributed charges of -2.00 µC and 3.50 µC. They are released from rest when their centers are separated by 1.00 m.
(a) How fast will each be moving when they collide? (Hint: Consider conservation of energy and of linear momentum.