Question

Can someone explain how to get these answers? ( Velocity/Acceleration Question)

Need Explanation Please.

Answer 1

(a) Average acceleration `="Total change in velocity"/"Total time taken for the change"`
From the figure.
`v_(x_(t=0))=0\ ms^-1`
`v_(x_(t=10.0))=4\ ms^-1`

Inserting these values in above equation we get
Average acceleration `=(4-0)/(10.0-0)=4/10=0.4\ ms^-2`

(b) Maximum positive value of acceleration will occur when rate of change of velocity in increasing sense (slope of the velocity-time curve) is maximum positive. From the figure we see that this occurs at `t=3\ s` To find the value of this acceleration, draw a tangent to the curve at `t=3`. Calculate the slope of the curve using `"Rise"/"Run"`.

(c) Acceleration is zero when rate of change of velocity is zero. Or the velocity-time graph becomes parallel to the `x`-axis. This happens at `t=6\ s`

(d) Maximum negative value of acceleration will occur when rate of change of velocity in decreasing sense (slope of the velocity-time curve) is maximum negative. From the figure we see that this occurs at `t=8\ s` To find the value of this acceleration, draw a tangent to the curve at `t=8`. Calculate the slope of the curve using `"Rise"/"Run"`. In this case observe that rise is actually fall and is negative as value of velocity is falling. Therefore, slope will be negative.