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/s²) 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/s² position  m ?

t = 5.6 s

acceleration  m/s²   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/s²     position =11.51 m

t = 5.6 s

acceleration =-0.0105 m/s²   position = 32.77 m ?