The position of a particle moving along the x axis is given by x = 6.0t2 1.0t3, where x is in meters and t in seconds. What is the position of the particle when it achieves its maximum speed in the positive x direction? About the author Everleigh
Given that, x=9t 2 −t 3 We know that, v= dt dx The maximum speed is v=18t−3t 2 v maximum, at dt dv =0 dt dv =18−6t The time will be 0=18−6t t=3 sec The position of the particle at 3 sec Put the value of t in equation (I) x=9×9−27 x=81−27 x=54 m The position of the particle will be 54 m. Hence, Option A is correct Reply
Answer:
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Explanation:
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Given that,
x=9t
2
−t
3
We know that,
v=
dt
dx
The maximum speed is
v=18t−3t
2
v maximum, at
dt
dv
=0
dt
dv
=18−6t
The time will be
0=18−6t
t=3 sec
The position of the particle at 3 sec
Put the value of t in equation (I)
x=9×9−27
x=81−27
x=54 m
The position of the particle will be 54 m.
Hence, Option A is correct