Thank you for reporting, we will resolve it shortly
Q.
The resulting $a$ - $t$ graph for the given $v$ - $t$ graph is correctly represented in
Motion in a Straight Line
Solution:
Average acceleration for different time intervals is the slope of $v-t$ graph, which is as follows
For $0\, s -10\, s , \bar{a}=\frac{(24-0) ms ^{-1}}{(10-0) s }=2.4\, ms ^{-2}$
For $10\, s -18\, s , \bar{a}=\frac{(24-24) ms ^{-1}}{(18-10) s }=0\, ms ^{-2}$
For $18\, s -20\, s , \bar{a}=\frac{(0-24) ms ^{-1}}{(20-18) s }=-12\, ms ^{-2}$
So, the corresponding $a-t$ graph for the given $v$ - $t$ graph is as follows
This is shown correctly in graph.
