Question

A polo ball is hit with a mallet off the edge of a cliff. Its x- and y-coordinates as functions of time are given by x = 18.2t and y = 4.04t − 4.90t2, where x and y are in meters and t is in seconds. (Do not include units in your answer.)

(a) Write a vector expression for the ball's position as a function of time (in m), using the unit vectors î and ĵ. (Give the answer in terms of t.)

r = ________ m

By taking derivatives, do the following.

(b) Obtain the expression for the velocity vector v as a function of time (in m/s). (Give the answers in terms of t.)

v = __________ m/s

(c) Obtain the expression for the acceleration vector a as a function of time (in m/s2).

a = __________ m/s2

(d) Next, use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the polo ball at t = 3.30 s. (Assume the position is in m, the velocity is in m/s and the acceleration is in m/s2.)

r = ________ m

v = _________ m/s

a = _________ m/s2

Answer 1

, a) the vector expression is, n = (-2. ) + (4 s t -+-+” 6) The Velocity is first derivative of position vector vador - to [(18.2+)i +(4.04t - 4.9€); ] = 18.2 i +[4.04 – 204.984]j = 18.2 i +(4.04-9.84)j mis c) The acceleration is, 118.2 i +(4.04 -9.86); = 0i+.049.8) - = -9. 8ý misr d) The position is, 51 = (18.2)(3-3) +4.04) (3.3) – 49 )(3.3) 7); = 60.061 – 40.0299 =60i-40; m The Velocity is V = 18.21 -28.3j mis The acceleration a=-9. 8 ý mist is