starting from rest a car moves with Uniform acceleration till it attains a velocity of 72 km/h in 30 s then it moves with Uniform velocity for some time afterwards is retards and come to state of rest after covering a distance of 50 m if total distance covered in 950m.
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Answer:
=15sec
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
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