dv/dt = -1/t^2
v = (1/t) + c
3 = 1 + c
c = 2
v = (1/t) + 2
if t = 4
v = 2.25
v = (1/t) + c
3 = 1 + c
c = 2
v = (1/t) + 2
if t = 4
v = 2.25
Given: a = -1/t^2
Integrating the acceleration equation, we get:
v = ∫ a dt = ∫ (-1/t^2) dt
To solve this integral, we can rewrite -1/t^2 as -t^(-2):
v = ∫ -t^(-2) dt
Now, integrating -t^(-2) with respect to t, we get:
v = -(-1)/(1) * t^(-1) + C
v = t^(-1) + C
Given that the velocity is 3 m/s when t = 1, we can substitute these values into the velocity equation to solve for the value of C:
3 = (1)^(-1) + C
3 = 1 + C
C = 2
Thus, the velocity equation becomes:
v = t^(-1) + 2
Now, let's substitute t = 4 into the velocity equation to find the velocity when t = 4:
v = (4)^(-1) + 2
v = 1/4 + 2
v = (1 + 8)/4
v = 9/4 m/s
Therefore, the velocity when t = 4 is 9/4 m/s.
Integrating the acceleration function will give us the velocity function. To integrate -1/t², we use the power rule of integration, which states that integrating x^n with respect to x gives (1/(n+1))x^(n+1).
So, integrating -1/t² with respect to t gives:
∫(-1/t²) dt = -∫(1/t²) dt = -(∫t^(-2) dt)
Using the power rule of integration, we find:
= -(-1/t^1) + C = 1/t + C,
where C represents the constant of integration.
Now, we can use the given information that when t = 1 second, the particle has a velocity of 3 m/s. Plugging this into the velocity equation, we can find the value of the constant of integration (C).
So, when t = 1, v = 3:
1/1 + C = 3
1 + C = 3
C = 3 - 1 = 2
Therefore, the velocity function is given by v = 1/t + 2.
Now, to find the velocity when t = 4 seconds, we substitute t = 4 into the velocity equation:
v = 1/4 + 2
v = 1/4 + 8/4
v = 9/4
Therefore, when t = 4 seconds, the velocity of the particle is 9/4 m/s.