From the above velocity-time graph of a body, identify the interval of time in which the acceleration is maximum. The distance travelled by the body in this interval be Xm. Then what is `x/10` ?

Answer

`25.00`

Explanation

Highest possible acceleration is from B to C.
`therefore a_("max") =` Slope of BC curve `= (Deltav)/(Deltat)`
`= (40-10)/(40 - 30) = 3 ms^(-2)`
`s_("max") = (v^(2) - u^(2))/(2a_("max")) = (40^(2) - 10^(2))/(2 xx 3) = 250 m`