Answered: A ball is thrown upward at time t=13 s… | bartleby

Related questions

Question

A ball is thrown upward at time t=13 s , and its acceleration from time t=13 s to t=21 s is given by the function a(t)= 7/(t-12) m/s^2

Use 4-time intervals equal length to overestimate, and then underestimate, the relative velocity of the rising ball from time t=13 s to t=21 s

a) overestiamte

b)underestimate

Expert Solution

Step by step

Solved in 4 steps with 3 images

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

т An object is initially located at (3,4) m with a velocity of io = -5£ – 3ŷ. A constant acceleration of å = т 2£ + 3ŷ is applied to the object for an interval of 5 seconds. a) What is the velocity of the object after 5 seconds? (Answer: 5£ + 12ŷ) b) What is the position of the object after 5 seconds? (Answer: 3£ + 26.5ŷ m) c) What is the average velocity of the object over the 5 second interval? (Answer: 0£ + 4.5ŷ–) d) Plot the acceleration vs time for both x and y e) Plot the velocity vs time for both x and y f) Plot the position vs time for both x and y
The acceleration of an object (in m/s) is given by the function a(t) = 7 sin(t). The initial velocity of the object is v(0) = %3D - 11 m/s. %3D a) Find an equation v(t) for the object velocity v(t) = %3D b) Find the object's displacement (in meters) from time 0 to time 3 meters c) Find the total distance traveled by the object from time 0 to time 3 meters
Assume the velocity of a particle is given by v(t)=t^2 -3t-4, in cm/s i) Find the displacement in the first 5 seconds: cm. ii) Find the total distance traveled in the first 5 seconds: cm. iii) Find the general form of the position function: s(t)= the acceleration of the particle at t=5: cm/s^2
4) Listen Position of a particle moving along the x-axis varies in time according to the following expression 772N x(1) = (-2,0 ""):* + (3.0"): What is its velocity at t=2.0 s? O- 6.0 m/s O- 24 m/s O 2.0 m/s O 16 m/s O- 32 m/s O -2.0 m/s O- 4.0 m/s O-16 m/s
For 4.95 seconds, a particle moves in a straight line according to the position function f(t) = (t³ + 1)(5 – t) – 5 where t is measured in seconds and f in feet. You can visualise this motion and verify many of your answers using a graph. First attempt all the problems by hand, and then graph the position function to verify your answers. a) What is the position of the particle at t = 0, t = 1, t = 2, t = 4.95? b) Find the velocity of the particle at time t. What is the velocity of the particle at t = 0, t = 1, t = 2, t = 4.95? c) When is the particle at rest? When is the particle moving forward?
The position of a particle in millimeters is given by s = 112 - 23t+t² where t is in seconds. Plot the s-t and v-t relationships for the first 16 seconds. Determine the net displacement As during that interval and the total distance D traveled. By inspection of the s-t relationship, what conclusion can you reach regarding the acceleration?
Is it possible for a particle to cover a distance of 18m while it has a displacement of only 5m? Explain.
And object moves with constant Acceleration 3.00 m/s^2 and over a time interval reaches a final velocity of 10.6 m/s (A)If it's original velocity is 5.3 m/s what is it's displacement during the time interval (B)What is the distance it traveled during this interval (C)If it's initial velocity is -5.3 m/s one is it's displacement during this interval (D) What is the total distance it travels during the interval in parts (c)
Phys2425 3. If you throw one object upward with an initial speed of 4m/s at 1.5m above the ground, find a) the total time of flight for this free fall motion, b) the final velocity just before touching the ground, and c) the maximum height with respect to the ground. 28