(1) A car drives down a road in such a way that its velocity ( in m/s) at time t (seconds) is
v(t)=3t^(1/2)+4
Find the car's average velocity (in m/s) between t=2 and t=8
(2) A circular swimming pool has a diameter of 10 m, the sides are 4 m high, and the depth of the water is 3.5 m. (The acceleration due to gravity is 9.8 m/s^2 and the density of water is 1000 kg/m^3.)
How much work (in Joules) is required to:
(a) pump all of the water over the side?
(b) pump all of the water out of an outlet 2 m over the side?
v(t)=3t^(1/2)+4
Find the car's average velocity (in m/s) between t=2 and t=8
(2) A circular swimming pool has a diameter of 10 m, the sides are 4 m high, and the depth of the water is 3.5 m. (The acceleration due to gravity is 9.8 m/s^2 and the density of water is 1000 kg/m^3.)
How much work (in Joules) is required to:
(a) pump all of the water over the side?
(b) pump all of the water out of an outlet 2 m over the side?
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Integrate v(t) with the limits t=2 and t=8 to find the displacement between those times:
8 ........................................… 8
∫ 3t^(1/2) + 4 dt = [2t^(3/2) + 4t] = [2*8^(3/2) + 4*8] - [2*2^(3/2) + 4*2] ~= 63.60 m
2 ........................................… 2
speed = distance / time = 63.60 / 6 = 10.6 m/s
For 2, you can just realise that the gravitational potential energy of the water is the same as if all the water were concentrated at 1.75 m above the bottom of the pool. Therefore:
GPE = mgh = 274889.36 * 9.8 * 1.75 = 4,714,352.48 J
You need to raise the water to a) 4m and b) 6m.
a) GPE required = 10,775,662.80
Work required = GPE required - GPE = 6.06 MJ
b) GPE required = 16,163,494.20
Work required = GPE required - GPE = 11.45 MJ
8 ........................................… 8
∫ 3t^(1/2) + 4 dt = [2t^(3/2) + 4t] = [2*8^(3/2) + 4*8] - [2*2^(3/2) + 4*2] ~= 63.60 m
2 ........................................… 2
speed = distance / time = 63.60 / 6 = 10.6 m/s
For 2, you can just realise that the gravitational potential energy of the water is the same as if all the water were concentrated at 1.75 m above the bottom of the pool. Therefore:
GPE = mgh = 274889.36 * 9.8 * 1.75 = 4,714,352.48 J
You need to raise the water to a) 4m and b) 6m.
a) GPE required = 10,775,662.80
Work required = GPE required - GPE = 6.06 MJ
b) GPE required = 16,163,494.20
Work required = GPE required - GPE = 11.45 MJ
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x=vt (distance)
deltaX=x1-x1=v(8)*8-v(2)*2 (distance driven from t=2s to t=8s)
deltax=6.364m
v_avg=deltaX/deltaT=6.364/(8-2)=1.0606… m/s
m=density*Volume=1000*10^2*pi/4=25000*… [kg]
m=78540 kg
note center of gravity is in the center (due symmetry) so
y0=3.5/2+(4-3.5)=2.25m (height from center of mass of water to upper edge of pool wall)
y1=2-3.5/2=0.25m (height to lift water to 2m)
in general E=mgh
a)
E= m*g*y0 (note: E>0, meaning we need to put some work in to lift water over the edge)
b)
E= m*g*y1 (note: E>0, we need to pu energy to lift water to 2m port)
c)
assuming we want to drain water (opening at 0m)
E=mgh=m*g*(0-3.5/2)
(note: E<0, meaning we get energy by letting watter flow out opening at pool bottom, water can be used to do work)
deltaX=x1-x1=v(8)*8-v(2)*2 (distance driven from t=2s to t=8s)
deltax=6.364m
v_avg=deltaX/deltaT=6.364/(8-2)=1.0606… m/s
m=density*Volume=1000*10^2*pi/4=25000*… [kg]
m=78540 kg
note center of gravity is in the center (due symmetry) so
y0=3.5/2+(4-3.5)=2.25m (height from center of mass of water to upper edge of pool wall)
y1=2-3.5/2=0.25m (height to lift water to 2m)
in general E=mgh
a)
E= m*g*y0 (note: E>0, meaning we need to put some work in to lift water over the edge)
b)
E= m*g*y1 (note: E>0, we need to pu energy to lift water to 2m port)
c)
assuming we want to drain water (opening at 0m)
E=mgh=m*g*(0-3.5/2)
(note: E<0, meaning we get energy by letting watter flow out opening at pool bottom, water can be used to do work)