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Q7. A car moves along a horizontal straight line with an acceleration a = 6(s)1/3 m/s². When t = 2s its displacements = 27 m and its velocity v = 27 m/s. Calculate the velocity and acceleration of the point when t = 4 s. (Ans: 75m/s and 30 m/s²) Q8. A particle moves on a vertical line with an acceleration a = 2,(v)1/2 m/s² .When t = 2 s, its displacement s = 64/3 m and its velocity v = 16 m/s. Determine the displacement, velocity, and acceleration of the particle when t = 3 s. (Ans: 41.7m, 25m/s and 10m/s²) Q9. The acceleration of a point moving on a vertical line is given by the equation a= 12t -20. It is known that its displacement s = -10 m at time t = 0 and that its displacement s = 10 m at time t = 5 s. Derive the equation of its motion s(t). (Ans: s(t) = 2t³ -10t² + 4t -10)

          Q7. A car moves along a horizontal straight line with an acceleration a = 6(s)1/3 m/s². When t = 2s its displacements = 27 m and its velocity v = 27 m/s. Calculate the velocity and acceleration of the point when t = 4 s. (Ans: 75m/s and 30 m/s²)

Q8. A particle moves on a vertical line with an acceleration a = 2,(v)1/2 m/s² .When t = 2 s, its displacement s = 64/3 m and its velocity v = 16 m/s. Determine the displacement, velocity, and acceleration of the particle when t = 3 s. (Ans: 41.7m, 25m/s and 10m/s²)

Q9. The acceleration of a point moving on a vertical line is given by the equation a= 12t -20. It is known that its displacement s = -10 m at time t = 0 and that its displacement s = 10 m at time t = 5 s. Derive the equation of its motion s(t). (Ans: s(t) = 2t³ -10t² + 4t -10)

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Q7. A car moves along a horizontal straight line with an acceleration a = 6(s)1/3 m/s². When t = 2s its displacements = 27 m and its velocity v = 27 m/s. Calculate the velocity and acceleration of the point when t = 4 s. (Ans: 75m/s and 30 m/s²)

Q8. A particle moves on a vertical line with an acceleration a = 2,(v)1/2 m/s² .When t = 2 s, its displacement s = 64/3 m and its velocity v = 16 m/s. Determine the displacement, velocity, and acceleration of the particle when t = 3 s. (Ans: 41.7m, 25m/s and 10m/s²)

Q9. The acceleration of a point moving on a vertical line is given by the equation a= 12t -20. It is known that its displacement s = -10 m at time t = 0 and that its displacement s = 10 m at time t = 5 s. Derive the equation of its motion s(t). (Ans: s(t) = 2t³ -10t² + 4t -10)

Added by Bonnie M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Anand Jangid

08/09/2024

Step 1

The acceleration is constant, so we can use the equations of motion. The velocity at any time t is given by v = u + at, where u is the initial velocity. Here, u = 27 m/s, a = 6√(1/3) m/s^2, and t = 4s. So, v = 27 + 6√(1/3)*4 = 75 m/s. The acceleration remains the Show more…

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