The friction coefficient between an athlete's shoes and the ground is 0.90. Suppose a superman wears these shoes and and races for 50 m. There is no upper limit on his capacity of running at high speeds. (a) Find the minimum time he will have to take in completing the 50 m starting from rest. (b) Suppose he takes exactly this minimum time to complete the 50 m, what minimum time will he take to stop.Concept of Physics - 1 , HC VERMA , Chapter 6:Friction "
Answers
Answered by
45
Solution :
********************
To reach the minimum time he has to move with maximum possible acceleration.
Let the maximum acceleration be a.
ma-μR=0
ma=μmg
a=μg=0.9x10=9 m/s2
A) Initial velocity=u= o m/s
t=?
a=9m/s2
s=5m
From s= ut +1/2 at²
50=0+1/2 at²
t=100/9
t=10/3 sec=3.3 sec
b) AFTER MOVING 50m, velocity of the athelete given by :
V=u+at=0+9x 10/3
he has to stop in minimum time, so deceleration is
-a=-9 m/s2
R=mg
ma=μR
a=μg =9m/s2[deceleration]
u1=30m/s
v1=0 m/s]t=v1-u1/a
t=0-30/-a
t=-30/-9
t=10/3 sec=3.3 sec
********************
To reach the minimum time he has to move with maximum possible acceleration.
Let the maximum acceleration be a.
ma-μR=0
ma=μmg
a=μg=0.9x10=9 m/s2
A) Initial velocity=u= o m/s
t=?
a=9m/s2
s=5m
From s= ut +1/2 at²
50=0+1/2 at²
t=100/9
t=10/3 sec=3.3 sec
b) AFTER MOVING 50m, velocity of the athelete given by :
V=u+at=0+9x 10/3
he has to stop in minimum time, so deceleration is
-a=-9 m/s2
R=mg
ma=μR
a=μg =9m/s2[deceleration]
u1=30m/s
v1=0 m/s]t=v1-u1/a
t=0-30/-a
t=-30/-9
t=10/3 sec=3.3 sec
Answered by
13
Answer:
Explanation:
b) AFTER MOVING 50m, velocity of the athelete given by :
V=u+at=0+9x 10/3
he has to stop in minimum time, so deceleration is
-a=-9 m/s2
R=mg
ma=μR
a=μg =9m/s2[deceleration]
u1=30m/s
v1=0 m/s]t=v1-u1/a
t=0-30/-a
t=-30/-9
t=10/3 sec=3.3 sec
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