a particle travels in a straight line ,such that for a short time 2s<t<6s its motion is described by v=(4/a)m/s where a is in m/s^@. if v=6m/s when t=2s,determine the particle the particle's acceleration when t=3s
Your umpteen posts have been removed.
Once you write up YOUR THOUGHTS, please re-post, and someone here will be happy to comment.
To determine the particle's acceleration when t=3s, we can use the equation a=dv/dt, where a is the acceleration, v is the velocity, and t is the time.
Given that the particle's motion for the time interval 2s<t<6s is described by v=(4/a) m/s, and v=6 m/s when t=2s, we can substitute these values into the equation.
At t=2s, v=6 m/s:
6 = (4/a)
To find the value of a, we can solve this equation for a.
Multiply both sides of the equation by a:
6a = 4
Divide both sides of the equation by 6:
a = 4/6
a = 2/3 m/s²
So, when t=2s, the particle's acceleration is 2/3 m/s².
Now, to find the particle's acceleration when t=3s, we need to differentiate the equation v=(4/a) with respect to t.
Differentiating both sides gives us dv/dt = -4/a² da/dt.
We already know dv/dt (velocity's rate of change with respect to time). So, let's substitute these values into the equation.
dv/dt = -4/a² da/dt
At t=2s, dv/dt = 6 m/s:
6 = -4/(2/3)² da/dt
Simplify the equation:
6 = -4/(4/9) da/dt
6 = -9 da/dt
To find the value of da/dt, we can solve this equation for da/dt.
Divide both sides of the equation by -9:
da/dt = 6/-9
da/dt = -2/3 m/s²
So, when t=3s, the particle's acceleration is -2/3 m/s². Therefore, the particle's acceleration when t=3s is -2/3 m/s².