Question

The position of a particle moving along the x axis depends on the time according to the equation x = ct3 - bt6, where x is in meters and t in seconds. Let c and b have numerical values 2.9 m/s3 and 2.0 m/s6, respectively. From t = 0.0 s to t = 1.3 s, (a) what is the displacement of the particle? Find its velocity at times (b) 1.0 s, (c) 2.0 s, (d) 3.0 s, and (e) 4.0 s. Find its acceleration at (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s.

Answer 1

Given the equation of the particle, we know that:

`x=(2.9(m)/(s^(3) ) )*t^(3) -(2(m)/(s^(6) ) )*t^(6)`

a) In
`t=0.0s` we evaluate
`t` in the former equation:

`x_(1) =(2.9(m)/(s^(3) ) )*(0)^(3) -(2(m)/(s^(6) ) )*(0)^(6)=0m`

In
`t=1.3s`

`x_(2) =(2.9(m)/(s^(3) ) )*(1.3s)^(3) -(2(m)/(s^(6) ) )*(1.3s)^(6)=-3,28232m`

a) So, we know that the displacement of te particle is given by:

`x=x_(2)-x_(1) =-3.28232m-0m=-3.28232m`

To find it's velocity, we need to derivate the equation of position by the formula:

`v_(x) =(dx)/(dt) =3ct^(2)-6bt^(5) =(8.7(m)/(s^(3) ))t^(2) -(12(m)/(s^(6) ) )t^(5)`

And evaluate this expression at each specified t:

b)
`v(1s)_(x) =(8.7(m)/(s^(3) ))(1s)^(2) -(12(m)/(s^(6) ) )(1s)^(5)=-3.3(m)/(s)`

c)
`v(2s)_(x) =(8.7(m)/(s^(3) ))(2s)^(2) -(12(m)/(s^(6) ) )(2s)^(5)=-349.2(m)/(s)`

d)
`v(3s)_(x) =(8.7(m)/(s^(3) ))(3s)^(2) -(12(m)/(s^(6) ) )(3s)^(5)=-2837.7(m)/(s)`

e)
`v(4s)_(x) =(8.7(m)/(s^(3) ))(4s)^(2) -(12(m)/(s^(6) ) )(4s)^(5)=-12148.8(m)/(s)`

To find it's acceleration, we need to derivate the equation of velocity by the formula:

`a_(x) =(dv)/(dt) =(17.4(m)/(s^(3) ) )t-(60(m)/(s^(6) ) )t^(4)`

And evaluate this expression at each specified t:

f)
`a(1s)_(x) =(17.4(m)/(s^(3) ) )(1s)-(60(m)/(s^(6) ) )(1s)^(4)=-42.6(m)/(s^(2) )`

g)
`a(2s)_(x) =(17.4(m)/(s^(3) ) )(2s)-(60(m)/(s^(6) ) )(2s)^(4)=-925.2(m)/(s^(2) )`

h)
`a(3s)_(x) =(17.4(m)/(s^(3) ) )(3s)-(60(m)/(s^(6) ) )(3s)^(4)=-4807.8(m)/(s^(2) )`

i)
`a(4s)_(x) =(17.4(m)/(s^(3) ) )(4s)-(60(m)/(s^(6) ) )(4s)^(4)=-15290.4(m)/(s^(2) )`