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A vehicle applies brake and skids through a distance of 60 m, before colliding with another parked vehicle, the weight of which is 75% of former. If distance travelled by both vehicle after collision is 15 m before stopping, then the initial speed (in m/sec) of moving vehicle was
[Assume collision is perfectly plastic, brake efficiency (η) = 100% and coefficient of friction (f) = 0.55]
- a)33.81
- b)47.25
- c)59.20
- d)64.80
Correct answer is option 'A'. Can you explain this answer?
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A vehicle applies brake and skids through a distance of 60 m, before ...
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A vehicle applies brake and skids through a distance of 60 m, before ...
Given:
Distance skidded before collision, s1 = 60 m
Weight of parked vehicle, w1 = 0.75w2
Distance travelled after collision, s2 = 15 m
Brake efficiency, η = 100%
Coefficient of friction, f = 0.55
To find:
Initial speed of moving vehicle, v1
Formula used:
v^2 = u^2 + 2as
where,
v = final velocity
u = initial velocity
a = acceleration
s = distance
f = coefficient of friction
g = acceleration due to gravity (9.8 m/s^2)
Calculation:
Let the weight of the moving vehicle be w2.
The force with which the brakes are applied is given by:
F = w2g
The force of friction is given by:
Ff = fF
= fw2g
The deceleration of the vehicle is given by:
a = Ff / w2
= fw2g / w2
= fgf
The initial velocity of the vehicle can be found using the formula:
v^2 = u^2 + 2as
Before collision:
s1 = (v1^2) / 2a
= (v1^2) / 2fgf
After collision:
s2 = (v2^2) / 2a
= (v2^2) / 2fgf
The final velocity of both the vehicles after the collision will be the same and can be found using the principle of conservation of momentum.
Let the velocity of both the vehicles after collision be v3.
The momentum of the moving vehicle before collision is given by:
p1 = w1v1
The momentum of the parked vehicle before collision is given by:
p2 = w2v2
The total momentum before collision is given by:
p1 + p2 = (w1v1) + (w2v2)
The total momentum after collision is given by:
p3 = (w1 + w2)v3
Using the principle of conservation of momentum, we get:
p1 + p2 = p3
(w1v1) + (w2v2) = (w1 + w2)v3
v3 = [(w1v1) + (w2v2)] / (w1 + w2)
Substituting the values of w1 and w2, we get:
v3 = [0.75w2v1 + w2v2] / (0.75w2 + w2)
= (0.75v1 + v2) / 1.75
Substituting the values of s1, s2, a and v3 in the formula v^2 = u^2 + 2as, we get:
(v3^2) = u^2 + 2as1 + 2as2
((0.75v1 + v2) / 1.75)^2 = u^2 + 2(fgf)(60) + 2(fgf)(15)
(0.75v1 + v2)^2 = (u^2) + 2(0.55)(9.8)(75)(60 + 15)
0.5625(v1^2) + 1.5v1v2 + (
Distance skidded before collision, s1 = 60 m
Weight of parked vehicle, w1 = 0.75w2
Distance travelled after collision, s2 = 15 m
Brake efficiency, η = 100%
Coefficient of friction, f = 0.55
To find:
Initial speed of moving vehicle, v1
Formula used:
v^2 = u^2 + 2as
where,
v = final velocity
u = initial velocity
a = acceleration
s = distance
f = coefficient of friction
g = acceleration due to gravity (9.8 m/s^2)
Calculation:
Let the weight of the moving vehicle be w2.
The force with which the brakes are applied is given by:
F = w2g
The force of friction is given by:
Ff = fF
= fw2g
The deceleration of the vehicle is given by:
a = Ff / w2
= fw2g / w2
= fgf
The initial velocity of the vehicle can be found using the formula:
v^2 = u^2 + 2as
Before collision:
s1 = (v1^2) / 2a
= (v1^2) / 2fgf
After collision:
s2 = (v2^2) / 2a
= (v2^2) / 2fgf
The final velocity of both the vehicles after the collision will be the same and can be found using the principle of conservation of momentum.
Let the velocity of both the vehicles after collision be v3.
The momentum of the moving vehicle before collision is given by:
p1 = w1v1
The momentum of the parked vehicle before collision is given by:
p2 = w2v2
The total momentum before collision is given by:
p1 + p2 = (w1v1) + (w2v2)
The total momentum after collision is given by:
p3 = (w1 + w2)v3
Using the principle of conservation of momentum, we get:
p1 + p2 = p3
(w1v1) + (w2v2) = (w1 + w2)v3
v3 = [(w1v1) + (w2v2)] / (w1 + w2)
Substituting the values of w1 and w2, we get:
v3 = [0.75w2v1 + w2v2] / (0.75w2 + w2)
= (0.75v1 + v2) / 1.75
Substituting the values of s1, s2, a and v3 in the formula v^2 = u^2 + 2as, we get:
(v3^2) = u^2 + 2as1 + 2as2
((0.75v1 + v2) / 1.75)^2 = u^2 + 2(fgf)(60) + 2(fgf)(15)
(0.75v1 + v2)^2 = (u^2) + 2(0.55)(9.8)(75)(60 + 15)
0.5625(v1^2) + 1.5v1v2 + (
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A vehicle applies brake and skids through a distance of 60 m, before colliding with another parked vehicle, the weight of which is 75% of former. If distance travelled by both vehicle after collision is 15 m before stopping, then the initial speed (in m/sec) of moving vehicle was[Assume collision is perfectly plastic, brake efficiency (η) = 100% and coefficient of friction (f) = 0.55]a)33.81b)47.25c)59.20d)64.80Correct answer is option 'A'. Can you explain this answer?
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A vehicle applies brake and skids through a distance of 60 m, before colliding with another parked vehicle, the weight of which is 75% of former. If distance travelled by both vehicle after collision is 15 m before stopping, then the initial speed (in m/sec) of moving vehicle was[Assume collision is perfectly plastic, brake efficiency (η) = 100% and coefficient of friction (f) = 0.55]a)33.81b)47.25c)59.20d)64.80Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A vehicle applies brake and skids through a distance of 60 m, before colliding with another parked vehicle, the weight of which is 75% of former. If distance travelled by both vehicle after collision is 15 m before stopping, then the initial speed (in m/sec) of moving vehicle was[Assume collision is perfectly plastic, brake efficiency (η) = 100% and coefficient of friction (f) = 0.55]a)33.81b)47.25c)59.20d)64.80Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vehicle applies brake and skids through a distance of 60 m, before colliding with another parked vehicle, the weight of which is 75% of former. If distance travelled by both vehicle after collision is 15 m before stopping, then the initial speed (in m/sec) of moving vehicle was[Assume collision is perfectly plastic, brake efficiency (η) = 100% and coefficient of friction (f) = 0.55]a)33.81b)47.25c)59.20d)64.80Correct answer is option 'A'. Can you explain this answer?.
A vehicle applies brake and skids through a distance of 60 m, before colliding with another parked vehicle, the weight of which is 75% of former. If distance travelled by both vehicle after collision is 15 m before stopping, then the initial speed (in m/sec) of moving vehicle was[Assume collision is perfectly plastic, brake efficiency (η) = 100% and coefficient of friction (f) = 0.55]a)33.81b)47.25c)59.20d)64.80Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A vehicle applies brake and skids through a distance of 60 m, before colliding with another parked vehicle, the weight of which is 75% of former. If distance travelled by both vehicle after collision is 15 m before stopping, then the initial speed (in m/sec) of moving vehicle was[Assume collision is perfectly plastic, brake efficiency (η) = 100% and coefficient of friction (f) = 0.55]a)33.81b)47.25c)59.20d)64.80Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A vehicle applies brake and skids through a distance of 60 m, before colliding with another parked vehicle, the weight of which is 75% of former. If distance travelled by both vehicle after collision is 15 m before stopping, then the initial speed (in m/sec) of moving vehicle was[Assume collision is perfectly plastic, brake efficiency (η) = 100% and coefficient of friction (f) = 0.55]a)33.81b)47.25c)59.20d)64.80Correct answer is option 'A'. Can you explain this answer?.
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