The three notches in the graph occur where the driver changed gears. Describe the changes in velocity and acceleration of the ca
r while in first gear. Is the acceleration just before a gear change larger or smaller than the acceleration just after the change? Explain your answer
<h3>Answer : </h3><h3 /><h3>A ) The larger gear can be moved by applying a relatively small force on the smaller gear.</h3><h3>B )Theforceappliedonthesmallergearistransmittedwithoutanylosstothelargergear.</h3><h3 /><h3>C)thedirectionofmotioncanbechangedwithoutchangingthedirectionoftheappliedforce.</h3>
To determine the speed relative to the ground, since the ground is our reference frame, it would be v = 0, for the kid on the skateboard, you would need to take into account the speed that he/she is going and the speed of the keys thrown at.
Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2 Time to height Tr = Vo / g Tr = (5.89 m/s) / (9.81 m/s^2) Tr = 0.60 s Max height achieved is: H = Vo^2 / [2g] H = (5.89 )^2 / [ 2 * (9.81) ] H = (34.69) / [19.62] H = 1.77 m It falls that distance, minus Andrew's catch distance: h = H - Y h = (1.77 m) - (1.27 m) h = 0.5 m Time to descend is therefore: Tf = √ { [2h] / g ] Tf = √ { [ 2 * (0.5 m) ] / (9.81 m/s^2) } Tf = √ { [ 1.0 m ] / (9.81 m/s^2) } Tf = √ { 0.102 s^2 } Tf = 0.32 s Total time is rise plus fall therefore: Tt = Tr + Tf Tt = (0.60 s) + (0.32 s) Tt = 0.92 s (ANSWER)
