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 need to integrate the velocity function with respect to time. The integral of velocity gives us the location or displacement.
Given: v(t) = (-5.00 x 10⁷)t² + (3.00 x 10⁶)t
To integrate this function, we can use the power rule of integration:
∫(at^n)dt = (a/(n+1)) * t^(n+1) + C,
where a is a constant, n is the exponent, t is the variable, and C is the constant of integration.
Applying the power rule to the velocity function, we have:
∫[(-5.00 x 10⁷)t² + (3.00 x 10⁶)t]dt
= [(-5.00 x 10⁷)/(2+1)] * t^(2+1) + [(3.00 x 10⁶)/(1+1)] * t^(1+1) + C
= -1.67 x 10⁷t³ + 1.50 x 10⁶t² + C
Since the position is defined at t=0 as being at the origin, we can set C=0.
Now, we can find the position after 10 seconds by evaluating the integral at t=10:
x(10) = -1.67 x 10⁷(10)³ + 1.50 x 10⁶(10)²
= -1.67 x 10⁷ x 1000 + 1.50 x 10⁶ x 100
= -1.67 x 10¹⁰ + 1.50 x 10⁸
= -1.66 x 10¹⁰ (rounded to two significant figures)
Therefore, the location of the particle after 10 seconds is approximately -1.66 x 10¹⁰ units.