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A car is moving along a straight road with uniform acceleration . It passes through two points P and Q separated by a distance with velocity 30 m/s and 40m/s respectively . The velocity of the car mid way between P and Q is A) 25√2 B) 35√2?
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A car is moving along a straight road with uniform acceleration . It p...
At mid point the velocity =under root v2+u2/2=(40)2+(30)2/2 1600+900/2=2500/2=under root 1250=25root2
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A car is moving along a straight road with uniform acceleration . It p...
Calculation of Velocity Midway between P and Q
- Let the initial velocity of the car at point P be u m/s
- Let the final velocity of the car at point Q be v m/s
- Distance between points P and Q = d meters
- Acceleration of the car = a m/s²
Using the first equation of motion:
v² = u² + 2ad
Substitute the given values:
40² = 30² + 2ad
Solve for acceleration:
a = (40² - 30²) / (2d)
Using the second equation of motion:
v = u + at
Substitute the values of u, a, and t = d/2 (as we are finding the velocity midway between P and Q):
v(midway) = u + a(d/2)
Substitute the values of u and a:
v(midway) = 30 + [(40² - 30²) / (2d)] * (d/2)
Simplify the equation:
v(midway) = 30 + (900 / d)
Given that d = 20m:
v(midway) = 30 + (900 / 20)
v(midway) = 30 + 45
v(midway) = 75 m/s
Therefore, the velocity of the car midway between points P and Q is 75 m/s.
Answer: The velocity of the car midway between P and Q is 75 m/s.
- Let the initial velocity of the car at point P be u m/s
- Let the final velocity of the car at point Q be v m/s
- Distance between points P and Q = d meters
- Acceleration of the car = a m/s²
Using the first equation of motion:
v² = u² + 2ad
Substitute the given values:
40² = 30² + 2ad
Solve for acceleration:
a = (40² - 30²) / (2d)
Using the second equation of motion:
v = u + at
Substitute the values of u, a, and t = d/2 (as we are finding the velocity midway between P and Q):
v(midway) = u + a(d/2)
Substitute the values of u and a:
v(midway) = 30 + [(40² - 30²) / (2d)] * (d/2)
Simplify the equation:
v(midway) = 30 + (900 / d)
Given that d = 20m:
v(midway) = 30 + (900 / 20)
v(midway) = 30 + 45
v(midway) = 75 m/s
Therefore, the velocity of the car midway between points P and Q is 75 m/s.
Answer: The velocity of the car midway between P and Q is 75 m/s.
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A car is moving along a straight road with uniform acceleration . It passes through two points P and Q separated by a distance with velocity 30 m/s and 40m/s respectively . The velocity of the car mid way between P and Q is A) 25√2 B) 35√2?
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A car is moving along a straight road with uniform acceleration . It passes through two points P and Q separated by a distance with velocity 30 m/s and 40m/s respectively . The velocity of the car mid way between P and Q is A) 25√2 B) 35√2? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A car is moving along a straight road with uniform acceleration . It passes through two points P and Q separated by a distance with velocity 30 m/s and 40m/s respectively . The velocity of the car mid way between P and Q is A) 25√2 B) 35√2? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car is moving along a straight road with uniform acceleration . It passes through two points P and Q separated by a distance with velocity 30 m/s and 40m/s respectively . The velocity of the car mid way between P and Q is A) 25√2 B) 35√2?.
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