Question

A car’s velocity as a function of time is given by Vx (t) = α.t + β.t 2 , where α= 3m/s and β= 0.1m/s 3 . Calculate the average acceleration for the time interval
b) t= 5 to t = 10 s

Answer 1

The definition of average acceleration allows to find the result for the average acceleration in the given time interval is:

          $a_{average}= 1.5 \ \frac{m}{s^2}$

Instantaneous acceleration is defined as the derivative of velocity with respect to time.

           a =   $\frac{dv}{dt}$

Where a is the acceleration, v the velocity and t the time.

They indicate that the speed of the car is given by the relation.

          v = α t + β t²

With α = 3 m / s and β = 0.1 m / s³

Let's make  the derivative.

           a = α + 2β t

Let's substitute

            a = 3 + 2 0.1 t

Average acceleration is the change in velocity in the time interval.  

          $a_{average} = \frac{\Delta v}{\Delta t }$

Let's find the velocity at the indicated time.

For t = 5 s

         v₅ = 3 + 0.1 5²

         v₅ = 5.5 m / s

For t = 10 s

          v₁₀ = 3 + 0.1 10²

          v₁₀ = 13 m / s

Let's calculate the average acceleration.

           $a_{average} = \frac{13 - 5.5 }{ 10 - 5 }$

           $a_{average}= 1.5 \ m/s^2$

In conclusion using the definition of mean acceleration we can find the result for the mean acceleration in the given time interval is:

           $a_{average} =$  1.5 m / s²

Learn more here: brainly.com/question/20057878