Question:
A car moving with uniform acceleration covers the distance of 200 m in first 2 seconds and the distance of 220 m in next 4 seconds. The velocity of the car after 7 seconds is
A car moving with uniform acceleration covers the distance of 200 m in first 2 seconds and the distance of 220 m in next 4 seconds. The velocity of the car after 7 seconds is
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Use the equations of motion for uniform acceleration. Set up equations based on the given distances and time intervals. Solve these simultaneous equations to find the initial velocity and acceleration. Then, use these values to find the velocity at the required time.
Updated On: May 12, 2025
- \( 10 \) ms\(^{-1} \)
- \( 20 \) ms\(^{-1} \)
- \( 15 \) ms\(^{-1} \)
- \( 30 \) ms\(^{-1} \)
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The Correct Option is A
Solution and Explanation
Let the initial velocity of the car be \( u \) and the uniform acceleration be \( a \).
Using the equation of motion \( s = ut + \frac{1}{2}at^2 \): For the first 2 seconds, the distance covered is 200 m: \( 200 = u(2) + \frac{1}{2}a(2)^2 \) \( 200 = 2u + 2a \) \( 100 = u + a \) (Equation 1) In the first \( 2 + 4 = 6 \) seconds, the total distance covered is \( 200 + 220 = 420 \) m: \( 420 = u(6) + \frac{1}{2}a(6)^2 \) \( 420 = 6u + 18a \) \( 70 = u + 3a \) (Equation 2) Subtracting Equation 1 from Equation 2: \( 70 - 100 = (u + 3a) - (u + a) \) \( -30 = 2a \) \( a = -15 \) ms\(^{-2} \) Substitute the value of \( a \) into Equation 1: \( 100 = u + (-15) \) \( u = 100 + 15 = 115 \) ms\(^{-1} \) The velocity of the car after \( t \) seconds is given by \( v = u + at \).
We need to find the velocity after 7 seconds: \( v(7) = u + a(7) \) \( v(7) = 115 + (-15)(7) \) \( v(7) = 115 - 105 \) \( v(7) = 10 \) ms\(^{-1} \)
Using the equation of motion \( s = ut + \frac{1}{2}at^2 \): For the first 2 seconds, the distance covered is 200 m: \( 200 = u(2) + \frac{1}{2}a(2)^2 \) \( 200 = 2u + 2a \) \( 100 = u + a \) (Equation 1) In the first \( 2 + 4 = 6 \) seconds, the total distance covered is \( 200 + 220 = 420 \) m: \( 420 = u(6) + \frac{1}{2}a(6)^2 \) \( 420 = 6u + 18a \) \( 70 = u + 3a \) (Equation 2) Subtracting Equation 1 from Equation 2: \( 70 - 100 = (u + 3a) - (u + a) \) \( -30 = 2a \) \( a = -15 \) ms\(^{-2} \) Substitute the value of \( a \) into Equation 1: \( 100 = u + (-15) \) \( u = 100 + 15 = 115 \) ms\(^{-1} \) The velocity of the car after \( t \) seconds is given by \( v = u + at \).
We need to find the velocity after 7 seconds: \( v(7) = u + a(7) \) \( v(7) = 115 + (-15)(7) \) \( v(7) = 115 - 105 \) \( v(7) = 10 \) ms\(^{-1} \)
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