Q14. The position of an object is given as = 2
2t^− 4 + 10 . Find the time t when the object is
stationary
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2
Answer:
x=t
2
−4t+6
So, the change in velocity is:
dt
dx
=2t−4
Since velocity is changing,
At t=0,x
1
=6
At t=2,x
2
=2
So the magnitude of the distance traveled is:
6−2=4m
At t=3,x
3
=3
So distance traveled from t=2s to t=3s
x
3
−x
2
=3−2=1m
Thus the total distance is:
4+1=5m
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