For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation dv/dt = g — v 2 . 420. Derive the previous expression for v(t) by integrating d v g − v 2 = d t .
For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation dv/dt = g — v 2 . 420. Derive the previous expression for v(t) by integrating d v g − v 2 = d t .
For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation dv/dt = g — v2.
420. Derive the previous expression for v(t) by integrating
d
v
g
−
v
2
=
d
t
.
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
A spring is stretched a distance x from its original length.
The potential energy stored in the spring is directly proportional to x squared.
The potential energy stored in the spring when it is stretched 0.25m is 600 J.
Derive an equation for the potential energy stored in the spring.
Note: You must determine the value of the constant.
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.
Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)
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