Rectilinear Motion with Variable Acceleration - Problem 5 - Kinematics of Particles

TL;DR

Determining the position and acceleration of a car at t=3s based on the provided equation of velocity.

Transcript

hi friends we'll solve problem on variable acceleration let's see what is given in problem during a test the car moves in a straight line such that its velocity is defined by v equal to 0.3 in bracket 90 square plus 2t meters per second where t is in seconds determine the position and acceleration when t equal to 3 second some boundary condition is... Read More

Key Insights

• 🧘 The position of an object can be determined by integrating its velocity equation.
• ❓ The acceleration of an object can be determined by differentiating its velocity equation.
• 🧘 Boundary conditions, such as position at a particular time, can be used to find constants of integration.
• ⌛ Substituting a specific time value into the position and acceleration equations gives the values at that time.

Questions & Answers

Q: What is the equation of velocity given in the problem?

The equation of velocity is v = 0.39t^2 + 2t, where t is in seconds.

Q: How is the equation of position obtained from the equation of velocity?

The equation of position is obtained by integrating the equation of velocity. By substituting dx/dt for v, the equation x = 0.9t^3 - 0.3t^2 is obtained.

Q: How is the equation of acceleration obtained from the equation of velocity?

The equation of acceleration is obtained by differentiating the equation of velocity. By differentiating 0.39t^2 + 2t with respect to t, the equation a = 5.4t - 0.6 is obtained.

Q: What are the position and acceleration at t = 3 seconds?

At t = 3 seconds, the position of the car is 27 meters and the acceleration is 16.8 meters per second square.

Summary & Key Takeaways

• The problem involves a car moving in a straight line with a velocity equation given as v = 0.39t^2 + 2t.

• By integrating the velocity equation, the position equation x = 0.9t^3 - 0.3t^2 is obtained.

• By differentiating the velocity equation, the acceleration equation a = 5.4t - 0.6 is obtained.

• At t = 3 seconds, the car's position is 27 meters and its acceleration is 16.8 meters per second square.