The position of a particle is given by the equation of motion s(t)= 1/(1+t)where t is in sec and s in meters find the velocity.
v = ds / dt = s´(t) = 1/ ( 1 + t ) ^ 2
would the velocity be the same after 2 seconds and would the speed at 2 seconds = be 9 feet/sec
velocity(0)=1
velocity(2)=1/9 m/s
No, and No to your answers.
To find the velocity of the particle, we need to differentiate the equation of motion with respect to time (t). The derivative of a function represents the rate of change of that function.
In this case, we have the equation of motion as s(t) = 1/(1 + t). To find the velocity, we need to calculate ds/dt, which is the derivative of s with respect to t.
Let's differentiate s(t) with respect to t using the chain rule:
ds/dt = d(1/(1 + t))/dt
To differentiate 1/(1 + t), we use the power rule, which states that d(u^n)/dt = n*u^(n-1)*du/dt, where u represents a function of t.
In this case, u = 1 + t and n = -1, so we have:
ds/dt = d(1/(1 + t))/dt
= -1/(1 + t)^2 * d(1 + t)/dt
Now, differentiating 1 + t with respect to t gives us 1 since the derivative of t is 1:
ds/dt = -1/(1 + t)^2 * 1
= -1/(1 + t)^2
So the velocity of the particle is given by -1/(1 + t)^2.
Therefore, the velocity equation for the particle is v(t) = -1/(1 + t)^2.