Starting at `x=0`, a particle moves according to the graph of `v` vs `t` shown in . Sketch a staph of the instantaneoud acceleration `a` vs `t`, indicationg numerical values at significant points of the graph.
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Answer

N//S

Explanation

To find a, we apply the relation `a=d v//dt`, which is the sane as `a=Delta v//Delta` over time intervals in which `v` varies linearly with `t`. For the time interval between `t_(0)=0` and `t_(1) =2 s`, the slope of the graph is positive and constant, equal to
` a_(1) =(Delta v)/(Delta t) =((v_(1) -v_(0)))/((t_(1)-t_(0))) =((2ms^(1) -0))/(2 s-0) =1 ms^(-2)`
Between `t_(1)=2 s` and `t_(2)=4 s`, the slope is zero. So the acceleration is zero for this time interval.
Between `t_(2)=4 s` and `t_(3)=5 s`, the slopeis negative and constant, equal to
.
`a_(2) =(Delta v)/(Delta t) =((v_(3)-v_(2)))/((t_(3)-t_(2)))`
`=((0-2ms^(-1)))/(5s-4s) =-2 ms^(-2)`
is a graph of the `a_(1) =(Delta v)/(Delta r) =((v_(1)-v_(0)))/((t_(1)-t_(0)))`
`=((2ms^(2)-0))/(2s-0) =1 m s^(-2)`
is a graph of the instantaneous acceleration for the motion.