Question

he function ​s(t)equalsnegative t cubed plus 27 t minus 2 gives the distance from a starting point at time t of a particle moving along a line. Find the velocity and acceleration functions. Then find the velocity and acceleration at tequals0 and tequals4. Assume that time is measured in seconds and distance is measured in centimeters. Velocity will be in centimeters per second​ (cm/sec) and acceleration in centimeters per second per second ​(cm/secsquared​)

Answer 1

`v(0)=27\\ v(4)=75`

`a(0)=0\\a(4)=24`

Explanation:

The function of the distance is equal to:

`s(t)=t^(3)+27t-2`

If we have the distance function, the velocity function is the derivative of s(t). So, the velocity function v(t) is equal to:

`v(t)=s'(t)\\v(t)= 3t^(2)+27`

Then, the velocity at t=0 and t=4 are calculated as:

`v(0)=3(0)^(2)+27=27\\ v(4)=3(4)^2+27=75`

So, the velocity at 0 seconds is equal to 27 cm/sec and the velocity at 4 seconds is equal to 75 cm/sec

At the same way, the acceleration function is the derivative of the velocity function. So, the acceleration function a(t) is equal to:

`a(t)=v'(t)\\a(t)=6t`

Then, the acceleration at t=0 and t=4 is:

`a(0)=6(0)=0\\a(4)=6(4)=24`

Therefore, the acceleration at 0 seconds is equal to 0
`cm/sec^(2)` and the acceleration at 4 seconds is equal to 24
`cm/sec^(2)`.