The position of an object moving along the $x$-axis is given by $x = a + b t^2$,where $a = 8.5 \; m$,$b = 2.5 \; m s^{-2}$,and $t$ is measured in seconds. What is the average velocity between $t = 2.0 \; s$ and $t = 4.0 \; s$ (in $; m s^{-1}$)?

$A$ body covers the first one-third of the distance with a velocity of $20 \, m/s$,the second one-third with a velocity of $30 \, m/s$,and the last one-third with a velocity of $40 \, m/s$. The average velocity is nearly........$m/s$.

$A$ man walks on a straight road from his home to a market $2.5 \,km$ away with a speed of $5 \,km/h$. Finding the market closed,he instantly turns and walks back home with a speed of $7.5 \,km/h$. The average speed of the man over the interval of time $0$ to $40 \,min$ is equal to

The position of an object moving along the $X$-axis is given by $x = \alpha + \beta t^2$, where $\alpha$ and $\beta$ are constants with appropriate dimensions and $t$ is time in seconds. The average velocity between $t = 2 \ s$ and $t = 4 \ s$ is $12 \ m/s$. If $\alpha = 8 \ m$, then the value of $\beta$ is: (in $m/s^2$)

If a car travels $40 \%$ of the total distance with a speed $v_1$ and the remaining distance with a speed $v_2$,then the average speed of the car is:

Give the difference between average speed and average velocity.