For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation d v / d t = g − v 2 . 421. [T] Estimate how far a body has fallen in 12 seconds by finding the area underneath the curve of v ( t ) .
For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation d v / d t = g − v 2 . 421. [T] Estimate how far a body has fallen in 12 seconds by finding the area underneath the curve of v ( t ) .
A 69-kg base runner begins his slide into second base when he is moving at a speed of 4.6 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
(a) How much mechanical energy is lost due to friction acting on the runner? J(b) How far does he slide? m
A particle starts to move at Q and travels in a straight line with velocity v cms-1, where v = 2t-t2 and t is the time in seconds after leaving Q. The particle comes to rest at A.
Calculate:
a. the value of t at A.
b. the acceleration, a cms-2
c. the distance, s cm, from Q to A.
d. the total distance covered from t=o to t=4.
An object is moving with velocity v(t) = 6t+ 6 in m/sec. Find the distance traveled by the object between time t = 0 and t = 5
seconds.
Select one:
Oa. 45 m
O b. 30 m
c. 105 m
d. 180 m
e. 36 m
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Area Between The Curve Problem No 1 - Applications Of Definite Integration - Diploma Maths II; Author: Ekeeda;https://www.youtube.com/watch?v=q3ZU0GnGaxA;License: Standard YouTube License, CC-BY