Explanation:
Avg speed × total times = total distance
20×25=500
Since acceleration in the 1st interval is same as deceleration in the 3rd interval and deceleration is from the same velocity achieved after acceleration, t
1
=t
3
s
1
=
2
1
×5t
1
2
s
2
=(5t
1
)t
2
s
3
=(5t
1
)×t
3
−
2
1
×5×t
1
2
=(5t
1
)×t
1
−
2
1
×5×t
1
2
s
1
+s
2
+s
3
=5t
1
t
2
+5t
1
2
⇒5t
1
2
+5t
1
t
2
=500−−−−1
2t
1
+t
2
=25−−−−2
Solving (1 ) and (2) ⇒t
1
=5,t
2
=15sec
