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anonymous

  • one year ago

A car is stopped at a traffic light. It then travels along a straight road so that its velocity is given by the equation v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2 Determine the following a. The average velocity of the car from t=0s to t=10s b. The instantaneous velocity of the car at t=10s c. The average acceleration of the car at t=10s d. The instantaneous acceleration of the car at t=10.s

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  1. anonymous
    • one year ago
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    Hopefully someone can help! Thankyouuuuuu @PlasmaFuzer hey sir if possible/want you can help me? If that is okay.

  2. anonymous
    • one year ago
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    or anyone else thank you! God bless people.

  3. Astrophysics
    • one year ago
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    Do you know calculus?

  4. anonymous
    • one year ago
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    Yes as Astrophysics pointed out, this problem is made quite simple with knowledge of some simple calculus and the relation of acceleration, velocity, and position functions... If: \[v(t) = (7.20 \frac{ m }{ s^{2} })t - (0.720 \frac{ m }{ s^{3} })t^{2} = c_{1}t + c_{2}t^{2}\] Where: \[c_{1}= (7.20 \frac{ m }{ s^{2} }) \ \ \ \ \& \ \ \ \ c_{2}= - (0.720 \frac{ m }{ s^{3} }) \] Then the position function can be found by integrating this equation: \[x(t) = \frac{1}{2}c_{1}t^{2} + \frac{1}{3}c_{2}t^{3}+x_{0}\] where x_0, read x naught, is an arbitrary constant of integration interpreted here as the initial position of the car can be taken to be 0.00m). As for the problem, and since we are on the subject of calculus, part (d) is found by differentiating v(t) with respect to time: \[a(t)=\frac{dv(t)}{dt} = c_{1} + 2c_{2}t \] This equation now tells you the instantaneous acceleration of the car at any time t, just as the velocity equation tells you the instantaneous velocity of the car at any time t. Therefore, the answer to part (d): \[a(10.0s)=c_{1} + 2c_{2}(10.0s)=-7.2 \frac{m}{s^{2}} \] Pay attention to the sign of c_2, and check your units!! And the answer to part (b) is (compute yourself first without looking to make sure you did it right!): \[v(t) = c_{1}(10.0s) + c_{2}(10.0s)^{2}=-72 \frac{m}{s} \] Again the sign of c_2, and again I strongly encourage you to check units though these equations and calculations! Looking at the units and knowing the units of the important quantities (acceleration, velocity, and position) can often guide you through what to do in most algebraic problems and can give you a definite check on you post calculus solutions (as well as provide a sanity check after doing the actual calculus itself). Now for the easy ones part (a) and part (c). Since it just asks for an averages, you simply need to know the total distance traveled divided by the total time (Ill leave for you to solve, hint: use the position function) to find the average velocity. As for the average acceleration, you already know the initial velocity was 0 m/s (check it plug in t=0.00s into the velocity equation) and you calculated the final velocity at time t=10.0s. Therefore: \[v_{avg}(t) = \frac{(-72 \frac{m}{s})-(0.00 \frac{ m }{ s })}{(10.0s)-(0.00s)} =-7.2 \frac{m}{s^{2}} \] I no doubt there were probably simpler ways of coming about the averages, and I invite additional comments. However since we went to the work of deriving all these equations these were the first that jumped to mind and I figured I would use them.

  5. anonymous
    • one year ago
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    Hey

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