Question

The position of an object moving along a line is given by `p(t) = t - 3sin(( pi )/3t)`. What is the speed of the object at `t = 4`?

Answer 1

`p(t)=t-3sin(pi/3t)`
`t=0 => p(0)=0m`
`t=4 => p(4)=4-3sin(pi/3*4)=>`
`p(4)=4-3sin(pi+pi/3)` (1)
`sin(pi+t)=-sin(t)` (2)
(1)+(2)`=>` `p(4)=4-(3*(-)sin(pi/3))=>`
`p(4)=4+3*sqrt(3)/2`
`p(4)=(8+3sqrt(3))/2m`

Now it depends on the extra information given:

1.If the acceleration isn't constant:
Using the law of space for the varied linear uniform movement:
`d=V""_0*t+(a*t^2)/2`
where
`d` is the distance,`V""_0` is the initial speed,`a` is the acceleration and `t` is the time when the object is in position `d`.

`p(4)-p(0)=d`
Assuming that the initial speed of the object is `0m/s`
`(8+3sqrt(3))/2=0*4+(a*16)/2=>`
`a=(8+3sqrt(3))/16m/s^2`

Finally the speed of the object at t=4 is
`V=a*4=(8+3sqrt(3))/4m/s`

2.If the acceleration is constant:
With the law of linear uniform movement:
`p(4)=p(0)+V(t-t""_0)`
You will get:
`(8+3sqrt(3))/2=0+V*4=>`
`V=(8+3sqrt(3))/8m/s`