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Answer:
4m/s
Let v,t
1
,t
2
be the maximum velocity attained, time for which car accelerated and the time for which car retarded respectively
⇒t
1
=(v−u)/a=(v−0)/2=v/2
⇒t
2
=(v−u)/a=(0−v)/(−4)=v/4
Now it is given that,
⇒t
1
+t
2
=3
⇒v/2+v/4=3
⇒v=4m/s
Explanation:
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