What is the maximum distance that the particle had from its starting point?
A particle moves along a straight line. The position of the particle is described by the expression x= 5t - 20t^2. Is the particle accelerating? What is the value of the acceleration? Initial velocity? Starting point? What time does the particle return to its starting point? What is the maximum distance that the particle had from its starting point?
1 Answer | Add Yours
The position with time is given by,
`x = 5t-20t^2`
Therefore, velocity is given by,
`v = (dx)/(dt) = 5 - 40t`
The acceleration is given by,
`a = (dv)/(dt) = -40`
It is accelerating in the opposite direction and the value of it is 40.
The initial velocity, at x = 0
v = 5-40t
v = 5
Therefore the initial velocity is 5.
The starting point is at `t = 0`
`x = 5t-20t^2`
`x = 0`
The starting point is `x = 0` .
To find the time it returns to the start poin we should solve for t when `x = 0`
`t-4t^2 = 0`
`t(1-4t) = 0`
`t = 0` is the starting time.
Therefore, `1-4t = 0, t = 1/4`
At `t=1/4` the particle comes to starting point again.
To find the maximumx can achieve we can rearange the equation as below.
`x = 5t -20t^2`
`x = -5(4t^2-t)`
x = -20(t^2-1/4t)
`x = -20(t^2-1/4t+(1/8)^2-(1/8)^2)`
`x = -20((t-1/8)^2-1/64)`
`x = -20(t-1/8)^2+5/16`
At `t = 1/8, x = 5/16` . Therefore the maximum distance it can travel is `5/16`
We’ve answered 319,865 questions. We can answer yours, too.Ask a question