Question

The velocity of a particle moving along the *x*-axis
varies with time according to

*v*(*t*) = *A* +
*Bt*^{−1},

where

*A* = 7 m/s,

*B* = 0.33 m,

and

1.0 s ≤ *t* ≤ 8.0 s.

Determine the acceleration (in m/s^{2}) and position (in
m) of the particle at

*t* = 2.6 s

and

*t* = 5.6 s.

Assume that

*x*(*t* = 1 s) = 0.

*t* = 2.6 s

acceleration m/s^{2} position m
?

*t* = 5.6 s

acceleration m/s^{2
}position m ?

Answer #1

From the given question,

*v*(*t*) = *A* +
*Bt*^{−1},

where *A* = 7 m/s, *B* = 0.33 m, 1.0 s ≤
*t* ≤ 8.0 s.

a(t)= dv/dt = 0 - Bt^-2

=-0.33t^-2

a(2.6)=-0.33(2.6)^-2=**-0.0489**

a(5.6)=0.33(5.6)^-2=**-0.0105**

position(x)= =

x =At + B logt + C

when t=1, x=0

0=A(1)+ Blog(1) +C

0=7+0+c

c=-7

x =7t + 0.33logt -7

x(2.6)=7(2.6) + 0.33log(2.6) -7

**x(2.6)=11.51 m**

x(5.6)=7(5.6) + 0.33log(5.6) -7

**x(5.6)=32.77 m**

*t* = 2.6 s

acceleration = **-0.0489**
m/s^{2 } position =**11.51
m**

*t* = 5.6 s

acceleration =**-0.0105** m/s^{2
}position = **32.77 m** ?

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