A ball has coordinates x = t2 – 3t + 20 and y = 2 + 3t given in inches, wheret is in seconds. Find the magnitude of the velocity and acceleration of the ball when t = 3 s.
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- The vertical displacement (in m) of a plane is 1.5 t2 up; t is the time (in s) after takeoff. The plane releases a package 2.0 s after takeoff. What is the magnitude of the velocity (in m/s) when the packaged is released? Report your answer in the appropriate number of significant digits. Answer valueAjetliner, traveling northward, is landing with a speed of 68.2 m/s. Once the jet touches down, it has 750 m of runway in which to reduce its speed to 7.16 m/s. Compute the average acceleration (magnitude and direction) of the plane during landing (take the direction of the plane's motion as positive). Number i Units 4A particle moves along two dimensions based on the following position vector r= [2.0 m + (3.00 m/s)f ]i + [(3.0 m)t - (2.00 m/s2)rlj a. Find the distance it covered in the first minute. b. Find the general expression of the instantaneous acceleration of the particle.
- Nerve impulses in a human body travel at a speed of about 100 m/s. Suppose a person accidentally steps barefoot on a pebble. About how much time does it take the nerve impulse to travel from the foot to the brain (in s)? Assume the person is 2.00 m tall and the nerve impulse travels at uniform speed. s ?We are standing on the top of a 1040 feet tall building and launch a small object upward. The object's height, measured in feet, after t seconds is h(t) = 16t? + 128t + 1040. A) What is the object initial velocity? ft/second B) What is the highest point that the object reaches? feetMy question isn't how to solve the problem exactly. In fact, it's already been solved on this website. My question is about the acceleration. When I solve this problem myself, first I calculate the velocity by dividing 100m by 53s. I get 1.89m/s. Then I use that to find the acceleration using the equation vf = vi + at. That's 1.89/53 = 0.036m/s^2. That's not correct. The correct way to find the acceleration is to us the equation d = 1/2 at^2 and solve that way without taking the intermediate step of finding the velocity. Doing it that way, the acceleration is 0.0712m/s^2. My question is why you get a different result doing it the first way than you get doing it the second way.
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- The coordinates of a particle moving along a curve are x(t) = -2t2 +15 and y(t) = t2 -10t + 15 , m whent is in seconds. Calculate the magnitude of velocity and acceleration when t= 5 sec.A particle moves along the x-axis according to the equation x 2.00 + 3.00t - the acceleration of the particle. 1.00t2. where x is in meters and t is in seconds. At t=D3.00s, find5 Velocity of a particle changes from (3î + 4j) m/s to (Oi + 5j) m/s is 2 s. Find magnitude and direction of average acceleration.