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?
I don't understand your expression for velocity.
position= Integral velocity*dt
To find the location of the particle after 10 seconds, we can use the equation for displacement, which is the integral of velocity with respect to time.
The equation for velocity is given as v(t) = (-5.00 x 10⁷)t² + (3.00 x 10⁶)t.
To find the displacement in this case, we need to integrate the velocity function over the interval from 0 to 10 seconds:
∫ [(-5.00 x 10⁷)t² + (3.00 x 10⁶)t] dt
Applying the power rule of integration, we can find the integral term by term:
= - (5.00 x 10⁷) * (t³/3) + (3.00 x 10⁶) * (t²/2)
Evaluating this expression from 0 to 10 seconds will give us the displacement of the particle after 10 seconds:
[( - (5.00 x 10⁷) * (10³/3) + (3.00 x 10⁶) * (10²/2)] - [( - (5.00 x 10⁷) * (0³/3) + (3.00 x 10⁶) * (0²/2)]
Simplifying the expression:
= [ - (5.00 x 10⁷) * (1000/3) + (3.00 x 10⁶) * (100/2)] - [0]
= - (5.00 x 10⁷) * (1000/3) + (3.00 x 10⁶) * (100/2)
= - (5.00 x 10⁷) * (333.33) + (3.00 x 10⁶) * (50)
Finally, calculating the expression:
= - 1.67 x 10¹⁰ + 1.50 x 10⁸
= - 1.67 x 10¹⁰ + 1.50 x 10⁸
≈ -1.67 x 10¹⁰ meters
Therefore, the particle's location after 10 seconds is approximately -1.67 x 10¹⁰ meters from the origin.