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A BUS WAS MOVING WITH SPEED OF 54 KM/HR ON APPLING BRAKES IT STOPED IN 8 SEC .CALCULATE ACCELERATION AND THE DISTANCE TRAVELLED BY BUS BY USING V=U+AT AND V2-U2=2AS
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A BUS WAS MOVING WITH SPEED OF 54 KM/HR ON APPLING BRAKES IT STOPED IN...
Calculation of Acceleration:
Given, initial velocity (u) = 54 km/hr
Final velocity (v) = 0 km/hr (as the bus stopped)
Time taken (t) = 8 sec
We know that, acceleration (a) = (v-u)/t
Substituting the given values, we get,
a = (0-54)/(8*5/18) [converting km/hr to m/s]
a = -7.5 m/s^2
Therefore, the acceleration of the bus is -7.5 m/s^2 (negative sign indicates that the bus is decelerating).
Calculation of Distance Travelled:
We can use the second equation of motion, which is v^2-u^2=2as, where s is the distance travelled by the bus.
Given, initial velocity (u) = 54 km/hr
Final velocity (v) = 0 km/hr (as the bus stopped)
Acceleration (a) = -7.5 m/s^2
We need to convert the initial velocity and acceleration to m/s.
Initial velocity (u) = 54 * 5/18 = 15 m/s
Acceleration (a) = -7.5 m/s^2
Substituting the given values, we get,
0^2 - 15^2 = 2*(-7.5)*s
Solving for s, we get,
s = 112.5 m
Therefore, the distance travelled by the bus before coming to a stop is 112.5 m.
Given, initial velocity (u) = 54 km/hr
Final velocity (v) = 0 km/hr (as the bus stopped)
Time taken (t) = 8 sec
We know that, acceleration (a) = (v-u)/t
Substituting the given values, we get,
a = (0-54)/(8*5/18) [converting km/hr to m/s]
a = -7.5 m/s^2
Therefore, the acceleration of the bus is -7.5 m/s^2 (negative sign indicates that the bus is decelerating).
Calculation of Distance Travelled:
We can use the second equation of motion, which is v^2-u^2=2as, where s is the distance travelled by the bus.
Given, initial velocity (u) = 54 km/hr
Final velocity (v) = 0 km/hr (as the bus stopped)
Acceleration (a) = -7.5 m/s^2
We need to convert the initial velocity and acceleration to m/s.
Initial velocity (u) = 54 * 5/18 = 15 m/s
Acceleration (a) = -7.5 m/s^2
Substituting the given values, we get,
0^2 - 15^2 = 2*(-7.5)*s
Solving for s, we get,
s = 112.5 m
Therefore, the distance travelled by the bus before coming to a stop is 112.5 m.
Community Answer
A BUS WAS MOVING WITH SPEED OF 54 KM/HR ON APPLING BRAKES IT STOPED IN...
acceleration =14.06 m/s²
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A BUS WAS MOVING WITH SPEED OF 54 KM/HR ON APPLING BRAKES IT STOPED IN 8 SEC .CALCULATE ACCELERATION AND THE DISTANCE TRAVELLED BY BUS BY USING V=U+AT AND V2-U2=2AS
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A BUS WAS MOVING WITH SPEED OF 54 KM/HR ON APPLING BRAKES IT STOPED IN 8 SEC .CALCULATE ACCELERATION AND THE DISTANCE TRAVELLED BY BUS BY USING V=U+AT AND V2-U2=2AS for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A BUS WAS MOVING WITH SPEED OF 54 KM/HR ON APPLING BRAKES IT STOPED IN 8 SEC .CALCULATE ACCELERATION AND THE DISTANCE TRAVELLED BY BUS BY USING V=U+AT AND V2-U2=2AS covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A BUS WAS MOVING WITH SPEED OF 54 KM/HR ON APPLING BRAKES IT STOPED IN 8 SEC .CALCULATE ACCELERATION AND THE DISTANCE TRAVELLED BY BUS BY USING V=U+AT AND V2-U2=2AS.
A BUS WAS MOVING WITH SPEED OF 54 KM/HR ON APPLING BRAKES IT STOPED IN 8 SEC .CALCULATE ACCELERATION AND THE DISTANCE TRAVELLED BY BUS BY USING V=U+AT AND V2-U2=2AS for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about A BUS WAS MOVING WITH SPEED OF 54 KM/HR ON APPLING BRAKES IT STOPED IN 8 SEC .CALCULATE ACCELERATION AND THE DISTANCE TRAVELLED BY BUS BY USING V=U+AT AND V2-U2=2AS covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A BUS WAS MOVING WITH SPEED OF 54 KM/HR ON APPLING BRAKES IT STOPED IN 8 SEC .CALCULATE ACCELERATION AND THE DISTANCE TRAVELLED BY BUS BY USING V=U+AT AND V2-U2=2AS.
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