Question 1181995: 1.An object moves along a straight path whose distance from the reference point is given by the position function, d(t) = t2 + 4t + 12 m after t seconds. Find:
a. the position of the object after 1, 2, 3, 4, and 5 seconds.
b. the average velocity of the object from 3 seconds to 4 seconds.
c. the displacement from 1 second to 2 seconds.
d. the instantaneous velocity after 3 seconds and after 5 seconds.
e. the acceleration at t second/s.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! d(t)=t^2+4t+12
d(1)=1+4+12=17 m
d(2)=4+8+12=24 m
d(3)=9+12+12=33 m
d(4)=16+16+12=44 m
d(5)=25+20+12=57 m
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average velocity from 1 to 2
d(2)-d(1) divided by 2-1
This is 7 m/sec
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displacement is 7 meters
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instantaneous velocity
dv/dt=2t+4
when t=3, dv/dt=10 m/sec velocity; t=5; dv/dt=14 m/sec velocity.
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acceleration at the seconds is 2nd derivative
That is 2t, and that is the acceleration at t seconds.
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