If a ball is thrown into the air with a velocity of 52 ft/s, its height in feet t seconds later is given by y = 52t – 162. (a) Find the average velocity for the time period beginning when t = and lasting for each of the following. (i) 0.5 seconds 70 ft/s (ii) 0.1 seconds 69.5 X ft/s (iii) 0.05 seconds 68.5 x ft/s 0.01 seconds (iv) 70 x ft/s (b) Estimate the instantaneous velocity when t = 2. 70 X ft/s

The position of a particle is given by s = 0.23t³ -0.77t² -2.50t + 5.65, where s is in feet and the time t is in seconds. Plot the displacement, velocity, and acceleration as functions of time for the first 12 seconds of motion. After you have the plots, answer the questions as a check on your work. -+s, ft -1 0 1 2 3 Questions: When t = 1.7 sec, V= ft/sec, ft/sec² When t = 9.8 sec, v = ft/sec, ft/sec² The positive time at which the particle changes direction is i i a = i a = sec.

Q10. As shown in the image below, the ball is kicked from point A with the initial velocity VA = 14 m/s at an angle of = 37°. Determine the maximum height hit reaches. Please pay attention: the numbers may change since they are randomized. Your answer must include 2 places after the decimal point, and proper SI unit. Take g = 9.81 m/s². VA Ө Your Answer: XB Answer R B h units

The velocity of a particle is given by v = 39t2-93t+ 47, where v is in meters per second and t is in seconds. Plot the velocity v and acceleration a versus time for the first 3 seconds of motion and evaluate the velocity when a is zero. Make the plots and then answer the questions. Questions: When t = 0.5 s, V = 10.25 m/s, a= -54 m/s² When t = 1.3 s, V= -7.99 m/s, a = 8.4 m/s² When t = 2.3 s, V= 39.41 m/s, a = 86.64 m/s² When a = 0, V= i 58.25 m/s

In class, we used the 1967 gravity formula to derive the change in gravitational acceleration g with latitude, and we calculated the result at latitude o = 45°. Plot the rate of change of g with latitude () in units of mgal/km, from the equator to the pole (ø = 0° to ø = 90°). Be careful with the units; you will need to convert radians to km. (Recall that there are 27 radians in a full circle, 360°.) do

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15-63. The four balls each have the same mass m. If A and B are rolling forward with velocity v and strike C, determine the velocity of each ball after the first three collisions. Take e=0.5 between each ball.

Determine the magnitude of the velocity of boat A with respect to boat B. Two boats leave the shore at the same time and travel in the directions shown in (Figure 1). vA = 12 m/s and vg = 19 m/s Express your answer to three significant figures and include the appropriate units. HẢ ? VA/B = Value Units Submit Request Answer Part B Determine the direction angle of the velocity of boat A with respect to boat B, measured counterclockwise from the positive a axis. Express your answer using three significant figures. ΠVα ΑΣφ It vec ? Figure Submit Request Answer Part C How long after leaving the shore will the boats be 760 m apart? Express your answer using three significant figures. \30 Πνα ΑΣφ It vec ? 45° t =

A some instant in time a rocket experiences an acceleration due to gravity and thrust from the rocket engine of: and has a velocity of: ā= 5.657î -3.843ĵ m/s² v = 5000 + 2000 m/s. Determine: i) the acceleration due to the rocket engine; ii) a, and an; iii) the radius of curvature of the trajectory at this instant ANS. at = 3.825 m/s^2, an = 5.669 m/s^2, p=5,115,541 m

A police car is traveling at a velocity of 20.0 m/s due north, when a car zooms by at a constant velocity of 47.0 m/s due north. After a reaction time 0.800 s the policeman begins to pursue the speeder with an acceleration of 3.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder? Number i Units Vos A VOS North AVP AVOP AVOP Police car catches up with speeder ↑ ap Police car begins to accelerate - Speeder passes police car