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Answer :
(a) The average speed between t=1.70 s and t=3.40 s:Average speed is given by $\frac{total\ distance}{total\ time}$.The distance covered between t=1.70 s and t=3.40 s:
[tex]$$x_2-x_1=(2.90*3.40^2-2.00*3.40+3.00)-(2.90*1.70^2-2.00*1.70+3.00)=26.46-5.29=21.17m$$[/tex]
The total time taken:
[tex]$$t_2-t_1=3.40-1.70=1.70s$$[/tex]
Therefore, the average speed is given by:
[tex]$$\frac{21.17}{1.70}=12.45m/s$$[/tex]
Answer: 12.45m/s
(b) The instantaneous speed at t=1.70 s:Instantaneous speed is given by differentiating the distance-time function.
[tex]$$\frac{dx}{dt}=5.84t-2.00$$[/tex]
At $t=1.70s$,
[tex]$$\frac{dx}{dt}=5.84*1.70-2.00=7.11m/s$$[/tex]
The instantaneous speed at t=3.40 s:
[tex]$$\frac{dx}{dt}=5.84*3.40-2.00=17.32m/s$$[/tex]
Answer: 7.11 m/s and 17.32 m/s
(c) The average acceleration between t=1.70 s and t=3.40 s:Acceleration is given by:
[tex]$$a_{av}=\frac{\Delta v}{\Delta t}$$[/tex]
$$a_{av}=\frac{17.32-7.11}{1.70}=6.00m/s^2$$
Answer: 6.00m/s2
(d) The instantaneous acceleration at t=1.70 s:The acceleration is given by the slope of the speed-time graph, which is the derivative of the distance function.
[tex]$$a=\frac{d^2x}{dt^2}=5.84m/s^2$$[/tex]
The instantaneous acceleration at t=3.40 s:
[tex]$$a=\frac{d^2x}{dt^2}=5.84m/s^2$$[/tex]
Answer: 5.84 m/s2
(e) At what time is the object at rest?
The object is at rest when it has zero velocity. So let us find the value of t when v=0
[tex]$$5.84t-2.00=0$$[/tex]
$$t=0.34s$$
Therefore the object is at rest at [tex]$t=0.34s$[/tex]
Answer: 0.34 s.
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