A particle moves on a straight line. The velocity after t seconds is given by V= 3t^2 – 6t – 8.
The distance of the particle from the origin after one second is 1o meters. Calculate the distance of the particle from the origin after 2 seconds.
s = t^3-3t^2-8t+c
use s(1) = 10 to find c,
then evaluate s(2)
To calculate the distance of the particle from the origin after 2 seconds, we first need to find the position function, or displacement, of the particle.
Given that the velocity function is V = 3t^2 - 6t - 8, the position function can be obtained by integrating the velocity function.
To integrate the velocity function, we will use the power rule of integration, which states that the integral of t^n is (1/n+1)(t^(n+1)).
Integrating the velocity function, we get:
∫ (3t^2 - 6t - 8) dt
= (1/3)(t^3) - (3/2)(t^2) - 8t + C
Here, C is the constant of integration.
To find the constant of integration, we are given that the distance of the particle from the origin after one second is 10 meters. Therefore, plugging in t = 1 in the position function should give us 10 meters.
(1/3)(1^3) - (3/2)(1^2) - 8(1) + C = 10
(1/3) - (3/2) - 8 + C = 10
1/3 - 3/2 - 8 + C = 10
1/3 - 9/6 - 8 + C = 10
2/6 - 9/6 - 48/6 + C = 10
-55/6 + C = 10
C = 10 + 55/6
C = 60/6 + 55/6
C = 115/6
Now, we have the complete position function:
P(t) = (1/3)(t^3) - (3/2)(t^2) - 8t + 115/6
To calculate the distance of the particle from the origin after 2 seconds, we simply substitute t = 2 into the position function:
P(2) = (1/3)(2^3) - (3/2)(2^2) - 8(2) + 115/6
P(2) = (1/3)(8) - (3/2)(4) - 16 + 115/6
P(2) = 8/3 - 12/2 - 16 + 115/6
P(2) = 8/3 - 18/3 - 16 + 115/6
P(2) = -10/3 - 16 + 115/6
P(2) = (-10 - 48 + 115)/6
P(2) = 57/6
P(2) = 9.5 meters
Therefore, the distance of the particle from the origin after 2 seconds is 9.5 meters.
To calculate the distance of the particle from the origin after 2 seconds, we need to first find the position function by integrating the velocity function.
Given:
Velocity function: v = 3t^2 - 6t - 8
To find the position function, we integrate the velocity function with respect to time:
s = ∫(3t^2 - 6t - 8) dt
Integrating each term separately:
s = ∫(3t^2) dt - ∫(6t) dt - ∫(8) dt
Integrating each term:
s = t^3 - 3t^2 - 8t + C
Now, we need to find the constant C by using the given information. We know that the distance of the particle from the origin after one second is 10 meters.
Substituting t = 1 and s = 10 into the position function:
10 = (1)^3 - 3(1)^2 - 8(1) + C
10 = 1 - 3 - 8 + C
10 = -10 + C
C = 20
Now we have the position function:
s = t^3 - 3t^2 - 8t + 20
To find the distance of the particle from the origin after 2 seconds, we substitute t = 2 into the position function:
s = (2)^3 - 3(2)^2 - 8(2) + 20
s = 8 - 12 - 16 + 20
s = 0
Therefore, the distance of the particle from the origin after 2 seconds is 0 meters.