Respuesta :
Answer:
Our equation for the height is:
y(t) = 275 - 16*t^2.
a) To find the average velocity between two times, t1 and t2, (where t2 > t1) the equation is:
[tex]AV = \frac{y(t2) - y(t1)}{t2 - t1}[/tex]
Then:
i) t1 = 4s, t2 = 4s + 0.1s = 4.1s
The average velocity is:
[tex]AV = \frac{(275 - 16*4.1^2) - (275 - 16*4^2)}{4.1 - 4} = \frac{16(4^2 - 4.1^2)}{0.1} = -129.6[/tex]
And the units will be ft/s, so the average speed is:
-129.6 ft/s
The minus sign is because te pebble is falling down.
ii) t1 = 4s, t2 = 4s + 0.05s = 4.05s
The average velocity is:
[tex]AV = \frac{(275 - 16*4.05^2) - (275 - 16*4^2)}{4.05 - 4} = \frac{16(4^2 - 4.05^2)}{0.05} = -128.8[/tex]
So the average speed is -128.9 ft/s
iii) t1 = 4s, t2 = 4s + 0.01s = 4.01s
The average speed is:
[tex]AV = \frac{(275 - 16*4.01^2) - (275 - 16*4^2)}{4.01 - 4} = \frac{16(4^2 - 4.01^2)}{0.01} = -128.16[/tex]
The average speed is -128.16 ft/s.
b) The instantaneous velocity of the pebble after 4 seconds can be obtained by looking at the velocity equation, that is the derivative of the height equation.
v(t) = dy(t)/dt.
v(t) = -2*16*t + 0
Then the velocity at t = 4s is:
v(4s) = -32*4 = -128
The instantaneous velocity at t = 4s is -128 ft/s.
a). The average velocity of the pebble for the time period would be:
i). [tex]-129.6[/tex]
ii). [tex]-128.8[/tex]
iii). [tex]-128.16[/tex]
b). The instantaneous velocity of the pebble post 4 seconds would be:
[tex]-128[/tex]
a). We need to find the average velocity,
i). [tex]t1 = 4s, t2 = 4s + 0.1s = 4.1s[/tex]
Using [tex]y(t_{2}) - (t_{1})/(t_{2}) - (t_{1})[/tex]
by putting the values,
[tex]= y(4.1) - y(4) / 4.1 - 4[/tex]
[tex]= 275 - 16(4.1)^2 - 275 - 16(4)^2 / 0.1[/tex]
[tex]= 275 - 16*16.81 - 275 - 16(16) / 0.1[/tex]
[tex]= 6.04 - 19 / 0.1[/tex]
[tex]= -12.96 /0.1[/tex]
[tex]= -129.6[/tex]
ii). Now,
[tex]t_{2} = (4 + 0.05 = 4.05s)[/tex]
Using [tex]y(t_{2}) - (t_{1})/(t_{2}) - (t_{1})[/tex]
[tex]= y(4.05) - y(4) / 4.05 - 4[/tex]
[tex]= -6.4 / 0.05[/tex]
[tex]= -128.8[/tex]
iii). Now,
[tex]t_{2} = 4.01s[/tex]
Using [tex]y(t_{2}) - (t_{1})/(t_{2}) - (t_{1})[/tex]
[tex]= y(4.01) - y(4) / 4.01 - 4[/tex]
[tex]= -1.28 / 0.01[/tex]
[tex]= -128.16[/tex]
b). Instantaneous velocity
[tex]y = 275 - 16t^2[/tex]
[tex]dy/dx = -32t[/tex]
Considering [tex]t = 4[/tex]
∵ Instantaneous velocity post 4 seconds [tex]= -32(4)[/tex]
[tex]= -128[/tex]
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