An object of mass mis moving along x-axis such that its position xis proportional to jź where tis time, then magnitude of its acceleration is
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The position of a particle moving along x-axis is given by
x=(−2t
3
+3t
2
+5)m
Velocity, v=
dt
dx
=0
−6t
2
+6t=0. . . . . . .(1)
6t(−t+1)=0
−t+1=0
t=1sec
Acceleration, a=
dt
dv
a=−12t+6
a∣
t=1sec
=−12×1+6
a∣
t=1sec
=−6m/s
2
The correct option is C.
x=(−2t
3
+3t
2
+5)m
Velocity, v=
dt
dx
=0
−6t
2
+6t=0. . . . . . .(1)
6t(−t+1)=0
−t+1=0
t=1sec
Acceleration, a=
dt
dv
a=−12t+6
a∣
t=1sec
=−12×1+6
a∣
t=1sec
=−6m/s
2
The correct option is C.
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