# The position of a particle moving along a coordinate line is s=√(1+4t), with s in meters and t in seconds. Find the particle's velocity and acceleration at t=6 seconds. This question appears in...

The position of a particle moving along a coordinate line is s=√(1+4t), with s in meters and t in seconds.

Find the particle's velocity and acceleration at t=6 seconds. This question appears in the Chain Rule lesson of the chapter so that must be used if necessary. Thanks so much for your help!

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The velocity is the derivative of the position with respect to time. The acceleration is the derivative of the velocity with respect to time. In finding each of the derivatives, it is necessary to use the chain rule, however, it is a relatively simple version in this case.

For velocity:

`v=d/{dt}sqrt{1+4t}`

`=1/2(1+4t)^{-1/2}(4)` where the 4 is from the derivative of 1+4t

`=2/sqrt{1+4t}`

Now substitute with t=6 to get:

`v(6)=2/sqrt{1+24}=2/5` m/s

The acceleration is:

`a=d/{dt}2(1+4t)^{-1/2}`

`=-2/2(1+4t)^{-3/2}(4)`

`=-4/(1+4t)^{3/2}`

And again sub in t=6 to get:

`a(6)=-4/(1+24)^{3/2}`

`=-4/5^3`

`=-4/125` m/s^2

**The velocity is `2/5` m/s and the acceleration is `-4/125` m/s^2.**