body starts from rest accelerates uniformly at the rate of 10 m/s² and retards uniformly at the rate of 20 cm/s² Find the least time in which it can complete the journey of 5Km to the maximum value attained by the body is 72 km/h
Answers
Answer:
It will accelerate to max possible speed with the max possible acceleration and then it will
slow down with the max possible negative acceleration
So the uniform acceleration is
1 = 10 cm/s² ∙ [1 m / 100 cm] = 0.1 m/s²
The maximum speed is
= 72 km/h ∙ [1000 m / km] x [1 h / 3600 s] = 20 m/s
and max uniform slowing acceleration is
2 = −20 cm/s² ∙ [1m / 100 cm] = −0.2 m/s² .
The time to accelerate to max speed = t1, at max speed from t1 to t2, slowing from 20 m/s to
0 m/s from t2 to t3 and that means we have to find t3 because t3 is the least time to complete the
trip.
Finding of t1:
1 =
− 0
1
=
20
0.1
= 200 s
The distance is
1 =
11
2
2
=
0.1 ∙ 2002
2
= 2000 m
The time of retarding is
2 =
0 −
2
=
−20
−0.2
= 100 s
The distance when retarding is
2 = 2 +
22
2
2
= 20 ∙ 100 −
0.2 ∙ 1002
2
= 1000 m
The remain distance is
3 = − 1 − 2 = 5000 − 2000 − 1000 = 2000 m
The distance d3 will moving at maximum speed, and time t3 is
3 =
3
=
2000
20 = 100 s
Total time is
= 1 + 2 + 3 = 200 + 100 + 100 = 400 s
So the least time required to complete the trip is 400 s.
Answer: 400 s.
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