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Questions and Answers
What is the position of the particle at $t = 4$ seconds if it starts at $x_o = 2$ m?
What is the position of the particle at $t = 4$ seconds if it starts at $x_o = 2$ m?
What is the acceleration of the object at $t = 3$ seconds?
What is the acceleration of the object at $t = 3$ seconds?
What is the displacement of the car between $t = 1$ and $t = 3$ seconds?
What is the displacement of the car between $t = 1$ and $t = 3$ seconds?
If the initial velocity of the object is 5 m/s, what is its velocity after 5 seconds?
If the initial velocity of the object is 5 m/s, what is its velocity after 5 seconds?
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What is the value of the velocity function at $t = 2$ seconds for the particle moving with $v(t) = 3t^2 - 2t + 1$?
What is the value of the velocity function at $t = 2$ seconds for the particle moving with $v(t) = 3t^2 - 2t + 1$?
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Study Notes
Problem 1: Particle Motion with Velocity Function
- Given velocity function: ( v(t) = 3t^2 - 2t + 1 ) m/s.
- Initial position: ( x_0 = 2 ) m at ( t = 0 ).
- To find position ( x(t) ), integrate the velocity function ( v(t) ).
- Integration yields: ( x(t) = t^3 - t^2 + t + C ).
- Apply initial condition to solve for ( C ): ( C = 2 ).
- Therefore, position function is ( x(t) = t^3 - t^2 + t + 2 ).
- Calculate position at ( t = 4 ): ( x(4) = 4^3 - 4^2 + 4 + 2 = 74 ) m.
Problem 2: Object Motion with Acceleration Function
- Given acceleration function: ( a(t) = 6t - 2 ) m/s².
- Initial conditions: velocity ( v(0) = 5 ) m/s and position ( x(0) = 3 ) m.
- To find position, integrate acceleration to find velocity: ( v(t) = 3t^2 - 2t + C ) where ( C = 5 ).
- Velocity function becomes ( v(t) = 3t^2 - 2t + 5 ).
- Integrate velocity to find position: ( x(t) = t^3 - t^2 + 5t + D ) where ( D = 3 ).
- Position function is ( x(t) = t^3 - t^2 + 5t + 3 ).
- Calculate position at ( t = 5 ): ( x(5) = 5^3 - 5^2 + 25 + 3 = 143 ) m.
Problem 3: Car Displacement Calculation
- Velocity function for the car: ( v(t) = 2t^3 - 3t^2 + 4t ) m/s.
- To calculate displacement, integrate the velocity function over the interval ( [1, 3] ).
- Integration leads to ( s(t) = \frac{1}{2}t^4 - t^3 + 2t + C ).
- Determine displacement: Displacement ( \Delta s = x(3) - x(1) ) using evaluated functions.
- Calculate ( x(3) ) and ( x(1) ):
- ( x(3) = \frac{1}{2}(3^4) - (3^3) + 2(3) ).
- ( x(1) = \frac{1}{2}(1^4) - (1^3) + 2(1) ).
- Result is the total displacement over the given interval.
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Description
This quiz focuses on solving motion problems using integration. You will calculate the position of a particle and an object's position based on given velocity and acceleration functions. Test your understanding of motion concepts in physics.
