|
3. The position of a particle moving along the x axis varies in time according to the expression `x = 3t^2`, where `x` is in meters and `t` is in seconds. Evaluate its position
(a) at `t = 3.00 s` and
(b) at `3.00 s + ∆t`.
(c) Evaluate the limit of `∆x//∆t` as `∆t` approaches zero to find the velocity at `t = 3.00 s`.
3. The position of a particle moving along the x axis varies in time according to the expression x = 3t^2, where x is in meters and t is in seconds. Evaluate its position
(a) at t = 3.00 s and
(b) at 3.00 s + ∆t.
(c) Evaluate the limit of ∆x//∆t as ∆t approaches zero to find the velocity at t = 3.00 s. `f(x) = ax^2 => f'(x) = 2ax`
(a) `27.0 m`
(b) `27.0 + 18.0(∆t) + 3.00(∆t)^2 m`
(c) `18.0 m//s`
(a)
When `t = 3.00 s`
`x = 3(3.00)^2 = color(red)(27.0 m)`
(b)
When `t = 3.00s + ∆t`
`x = 3(3.00 + (∆t))^2 = 3(9.00 + 6.00(∆t) +(∆t)^2)`
`x = color(red)(27.0 + 18.0(∆t) + 3(∆t)^2 m)`
(c)
Solve by Algebra:
Here we set `x` form (a) as `x_i`
set `x` form (b) as `x_f`
`∆x = x_f - x_i`
`=> ∆x = 27.0 + 18.0(∆t) + 3(∆t)^2 - 27.0`
`=> ∆x = 18.0(∆t) + 3(∆t)^2 `
`(∆x)/(∆t) = (18.0(∆t) + 3(∆t)^2)/(∆t)`
`=18.0 +3∆t`
limit of `(∆x)/(∆t)` as `∆t` approaches zero write as:
`lim_(∆t->0)(18.0 +3∆t)`
when `∆t` approaches to `0` `3∆t = 0`
` lim_(∆t->0)(18.0 +3∆t) = 18.0 m//s`
Solve by calculus:
`x = 3t^2`
`=> v = (∆x)/(∆t) = 2•3t =6t`
at `t = 3.00s`
`v = 6(3.00) m//s = color(red)(18.0 m//s)`
|