Physics Motion Problems

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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?

  • 34 m (correct)
  • 20 m
  • 22 m
  • 16 m

What is the acceleration of the object at $t = 3$ seconds?

  • 10 m/s² (correct)
  • 8 m/s²
  • 14 m/s²
  • 16 m/s²

What is the displacement of the car between $t = 1$ and $t = 3$ seconds?

  • 12 m
  • 4 m
  • 8 m (correct)
  • 6 m

If the initial velocity of the object is 5 m/s, what is its velocity after 5 seconds?

<p>27 m/s (D)</p> Signup and view all the answers

What is the value of the velocity function at $t = 2$ seconds for the particle moving with $v(t) = 3t^2 - 2t + 1$?

<p>7 m/s (A)</p> Signup and view all the answers

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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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