Question

Velocity of a car as function of time is given by vx(t)=α+βt2, where α=3.00ms−1 and β=0.100ms−3. The average acceleration for the time interval t = 0 and t = 5s is

Answer 1

Answer:

The time interval (t) varies from one instance to another, on which the values of velocity and the average acceleration depend.

If t = 0, vx(t) = vx (0) = 0, thus the average acceleration is also calculated to be 0.

Answer 2

Answer:

0.5 $ms^{-2}$

Explanation:

         Formula :

                          Average acceleration $=\frac{v_2-v_1}{t_2-t_1}$

         $v_{x}(t)=\alpha +\beta t^{2}$

         Let time $t_{1}$ = 0s  and time $t_{2}$ = 5s

         $\alpha$ $=3.00ms^{-1}$            $\beta$ $= 0.100ms^{-3}$

         $v_{1}$(at t = 0s) = 3 + (0.1)($0^{2}$)  

     

         $v_{1}$(at t = 0s) = $3ms^{-1}$--------> Equation-1

         $v_{2}$(at t= 5s) = 3 + (0.1)($5^2$)    

          $v_{2}$(at t= 5s) = $5.5ms^-1$--------> Equation-2

           THEREFORE :

        Average acceleration $=\frac{v_2-v_1}{t_2-t_1}$

                                              $=\frac{5.5-3}{5-0}$

                                              $=\frac{2.5}{5}$

         Average acceleration $=0.5ms^{-2}$

                                        HOPE YOU UNDERSTOOD

                                                   THANK YOU