An antelope moving with constant acceleration covers the distance between two points 70.0
m apart in 7.00 s. Its speed as it passes the second point is 15.0 m/s. (a) What is its speed at the first
point? (b) What is its acceleration?
A car is stopped at a traffic light. It then travels along a straight road so that its distance from
the light is given by x(t) = bt2 – ct3, where b = 2.40m/s2 and c = 0.120m/s3. (a) Calculate the average
velocity of the car for the time interval t = 0 to t = 10.0 s. (b) Calculate the instantaneous velocity of the
car at t = 0, t = 5.0 s, and t = 10.0 s. (c) How long after starting from rest is the car again at rest?
ONe question per post, please. I helps you to put down what you think needs to be done, or what you have tried, that way, we can see where your thinking is going askew, and correct that.
avg velocity= (Vf+Vi)/2=distance/time
solve for Vi
Vf=Vi+at solve for a
Water flows (velocity initial=0) over a dam at the rate of 660 kg/s and falls vertically 81 m before striking the turbine blades. Calculate the rate at which mechanical energy is transfered to the turbine blades, assuming 55% efficiency.
(a) To find the antelope's speed at the first point, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity = 15.0 m/s
u = initial velocity (unknown)
a = acceleration (unknown)
s = distance traveled = 70.0 m
Rearranging the equation, we have:
u^2 = v^2 - 2as
Substituting the given values, we get:
u^2 = 15.0^2 - 2 * a * 70.0
Simplifying:
u^2 = 225.0 - 140.0a
Since we know the antelope is moving with constant acceleration, we can assume that the acceleration remains the same throughout the whole motion. Therefore, the antelope's acceleration at the first point will be the same as the acceleration when it passes the second point.
(b) To find the acceleration, we can use the formula:
s = ut + (1/2)at^2
Where:
s = distance traveled = 70.0 m
u = initial velocity (unknown)
t = time taken = 7.00 s
a = acceleration (unknown)
We need to rearrange the equation to isolate the acceleration:
a = (2s - ut^2) / t^2
Substituting the given values, we get:
a = (2 * 70.0 - u * 7.00^2) / 7.00^2
Simplifying:
a = (140.0 - 49.0u) / 49.0
For the second part of the question:
(a) To calculate the average velocity of the car from t=0 to t=10.0 s, we need to find the displacement and divide it by the time interval.
Displacement, Δx = x(t2) - x(t1)
Where:
x(t2) = bt2 - ct3 at t = 10.0 s
x(t1) = bt2 - ct3 at t = 0
Substituting the values:
Δx = x(t2) - x(t1)
Δx = (b * 10.0^2 - c * 10.0^3) - (b * 0^2 - c * 0^3)
Δx = 100b - 1000c
Average velocity, v_avg = Δx / Δt = Δx / (t2 - t1)
v_avg = (100b - 1000c) / (10.0 - 0)
(b) To calculate the instantaneous velocity at t = 0, t = 5.0 s, and t = 10.0 s, we can take the derivative of the position function x(t) with respect to time:
v(t) = d(x(t))/dt = d(bt^2 - ct^3)/dt
v(t) = 2bt - 3ct^2
Substituting the values, we get:
v(0) = 2b * 0 - 3c * 0^2 = 0
v(5.0) = 2b * 5.0 - 3c * 5.0^2
v(10.0) = 2b * 10.0 - 3c * 10.0^2
(c) To find the time it takes for the car to come to rest, we need to find when its velocity v(t) becomes zero. We set the equation v(t) = 0 and solve for t:
0 = 2bt - 3ct^2
2bt = 3ct^2
t = (2b)/(3c)
Substituting the given values of b and c, we can calculate the time it takes for the car to come to rest.
To solve these physics problems, we will need to apply the relevant equations and principles. I will guide you through the steps to find the answers:
1) Antelope moving with constant acceleration:
(a) To find the speed at the first point, we have to determine the initial velocity of the antelope. We are given the following information:
- Distance between two points (displacement): 70.0 m
- Time taken to cover this distance: 7.00 s
- Speed at the second point: 15.0 m/s
We can use the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Since the antelope is moving with constant acceleration, we can use the average velocity to find the initial velocity:
Average velocity = (Final velocity + Initial velocity) / 2
Using the given values, let's calculate the average velocity:
Average velocity = (15.0 m/s + u) / 2
We know that the average velocity is the displacement divided by the time taken:
Average velocity = Displacement / Time
Average velocity = 70.0 m / 7.00 s
Now we can equate the two expressions for average velocity and solve for the initial velocity:
(15.0 m/s + u) / 2 = 70.0 m / 7.00 s
Solving for u, the initial velocity:
15.0 m/s + u = 40.0 m/s
u = 40.0 m/s - 15.0 m/s
Therefore, the speed at the first point is:
u = 25.0 m/s
(b) Now, to find the acceleration, we can use the equation of motion v^2 = u^2 + 2as, where s is the displacement. Rearranging the equation to isolate acceleration, we have:
v^2 - u^2 = 2as
We are given:
- Final velocity (speed at the second point): v = 15.0 m/s
- Initial velocity (speed at the first point): u = 25.0 m/s
- Displacement: s = 70.0 m
Plug in the values into the equation and solve for acceleration (a):
15.0^2 - 25.0^2 = 2a(70.0)
a = (225.0 - 625.0) / (2 * 70.0)
Therefore, the acceleration is:
a = -8.93 m/s^2 (negative sign indicates deceleration)
2) Car stopped at a traffic light:
(a) To find the average velocity of the car from t = 0 to t = 10.0 s, we need to calculate the displacement during that time interval. The equation provided for the distance from the light as a function of time is x(t) = bt^2 - ct^3, where b = 2.40 m/s^2 and c = 0.120 m/s^3.
To find the average velocity, we divide the displacement by the time interval:
Average velocity = Displacement / Time interval
Substituting the given values into the equation for x(t):
Displacement = x(10.0 s) - x(0)
Displacement = (2.40 m/s^2 * (10.0 s)^2) - (0.120 m/s^3 * (10.0 s)^3)
Now we can calculate the displacement and divide it by the time interval (10.0 s) to find the average velocity.
(b) To calculate the instantaneous velocity at different times (t = 0 s, t = 5.0 s, and t = 10.0 s), we take the derivative of the position function x(t) with respect to time t. The derivative gives us the velocity function, which represents the instantaneous velocity at any given time.
To find the instantaneous velocity at t = 0 s, t = 5.0 s, and t = 10.0 s, substitute the respective values of t into the velocity function.
(c) To determine how long after starting from rest the car is again at rest, we need to find the time when the velocity function is zero. Set the velocity function equal to zero and solve for t. This will give us the time at which the car comes to rest again.
Remember to use the given values of b = 2.40 m/s^2 and c = 0.120 m/s^3 in all the calculations.
Please let me know if you need further assistance with any specific step in solving these problems.