The velocity of a particle is given by the relation v(t)=(-5.00 x 10⁷)t² = (3.00 x 1065)t. At time t=0, the particle is located at the origin. What is its location after 10 seconds?
To find the location of the particle after 10 seconds, we can integrate the velocity function with respect to time. The integral of velocity gives us the displacement or change in position.
The velocity function is given as v(t) = (-5.00 x 10⁷)t² + (3.00 x 10⁶)t.
To integrate, we need to find the antiderivative of the function. The antiderivative of t² is (1/3)t³, and the antiderivative of t is (1/2)t².
Using these antiderivatives, we can integrate the velocity function:
∫v(t) dt = ∫[(-5.00 x 10⁷)t² + (3.00 x 10⁶)t] dt
= (-5.00 x 10⁷)∫t² dt + (3.00 x 10⁶)∫t dt
= (-5.00 x 10⁷)(1/3)t³ + (3.00 x 10⁶)(1/2)t² + C
Here, C represents the constant of integration.
Now, we can substitute the limits of integration. Since we want to find the location after 10 seconds, we evaluate the integral from t = 0 to t = 10:
∫v(t) dt from 0 to 10 = [(-5.00 x 10⁷)(1/3)t³ + (3.00 x 10⁶)(1/2)t² + C] evaluated from 0 to 10
Plugging in the values:
[(-5.00 x 10⁷)(1/3)(10)³ + (3.00 x 10⁶)(1/2)(10)² + C] - [(-5.00 x 10⁷)(1/3)(0)³ + (3.00 x 10⁶)(1/2)(0)² + C]
Simplifying further:
[(-5.00 x 10⁷)(1/3)(1000) + (3.00 x 10⁶)(1/2)(100) + C] - [(-5.00 x 10⁷)(1/3)(0) + (3.00 x 10⁶)(1/2)(0) + C]
[(-5.00 x 10⁷)(1000/3) + (3.00 x 10⁶)(100/2) + C] - [0 + 0 + C]
Evaluating these values:
= (-5.00 x 10⁷)(1000/3) + (3.00 x 10⁶)(100/2) + C
= (-1.67 x 10¹⁷) + (1.50 x 10⁸) + C
Now, we need to determine the constant of integration, C. Since the particle is located at the origin (position 0) at t = 0, we can set the displacement at t = 0 equal to 0. This implies that C = 0.
Substituting C = 0:
= (-1.67 x 10¹⁷) + (1.50 x 10⁸) + 0
= -1.67 x 10¹⁷ + 1.50 x 10⁸
Therefore, the location of the particle after 10 seconds is approximately -1.67 x 10¹⁷ + 1.50 x 10⁸ units.