a motor bike running at 94km per hr and slowed at 54km per hr by the application of breaks over a distance of 40 m if the breaks applied with the same force then the total time in which bike comes to rest calculate the time for which breaks are aplied
Answers
Answer:
Time taken for changing velocity = 0.0005 s
time taken to come in rest = 0.0007s
Explanation:
here given that,
a motor bike running at 94km/h
here initial velocity of the motor bike
= 94 km/h
also given
It slowed at 54km per hr by the application of breaks over a distance of 40 m
here,
let the final velocity of the motor bike
be 54 km/h
given distance covered during changing the velocity = 40 m
= 40/1000 km
= 1/25 km
= 0.04 km
now we have,
initial velocity(u) = 94 km/h
final velocity(v) = 54 km/h
distance covered(s) = 0.04 km
so,
by the equation of the motion
v² = u² + 2as
where
a = acceleration
putting the values,
(54)² = (94)² + 2a(0.04)
2916 = 8836 + 0.08a
0.08a = 2916 - 8836
0.08a = -5920
a = -5920/0.08
a = -74000 km/h²
so,
acceleration(a) = -74000 km/h²
now,
by the equation of motion
v = u + at
putting the values,
54 = 94 + (-74000)t
-74000t = -54 - 94
-74000t = -40
t = -40/-74000
t = 0.0005 s
so,
Time taken for changing velocity
= 0.0005 s
now,
we have,
initial velocity (u) = 54 km/h
final velocity(v) = 0
acceleration(a) = -74000 m/s²
v = u + at
putting the values,
0 = 54 - 74000t
-74000t = -54
t = -54/-74000
t = 0.0007 s
so,
time taken to come in rest = 0.0007s
Answer:
Time taken for changing velocity
= 0.0005 s
time taken to come in rest = 0.0007s
Explanation:
given that,
initial velocity of a motor bike(u) = 94km/h
final velocity of the motor bike(v) = 54 km/h distance covered after applying brakes(s)
= 40 m
= 40/1000 km
= 1/25 km
= 0.04 km
v² = u² + 2as
(54)² = (94)² + 2a(0.04)
2916 = 8836 + 0.08a
0.08a = 2916 - 8836
0.08a = -5920
a = -5920/0.08
a = -74000 km/h²
acceleration(a) = -74000 km/h²
v = u + at
54 = 94 + (-74000)t
94 -74000t = 54
-74000t = -40
t = -40/-74000
t = 0.0005 s
Time taken to change velocity
= 0.0005 s
to find,
time taken to come in rest with the same retardation
initial velocity (u) = 54 km/h
final velocity(v) = 0
acceleration(a) = -74000 m/s²
v = u + at
0 = 54 - 74000t
-74000t = -54
t = -54/-74000
t = 0.0007 s
time taken to come in rest = 0.0007s
_____________________
NOTE:
NEGATIVE ACCELERATION SHOWS THE RETARDATION