Question

Can someone please help me with these questions???

Answer 1

Velocity at `t=1` is `-3`, speed is `3`, accelaration is `42` and jerk is `-230`

Explanation:

If euation of motion is given by `f(t)`, then

velocity is given by `f'(t)`, speed is `|f'(t)|`, accelaration is given by `f''(t)` and jerk is `f'''(x)`.

As `f(t)=4t^(-3)+27t^(1/3)+12`

velocity is given by `f'(t)=4xx(-3)t^(-4)+27xx1/3t^(-2/3)`

= `-12t^(-4)+9t^(-2/3)`

velocity at `t=1` is `f'(1)=-12+9=-3`

and speed at `t=1` is `|f'(1)|=|-3|=3`

As velocity is `f'(t)=-12t^(-4)+9t^(-2/3)`,

accelaration is given by its derivative i.e.

`f''(t)=-12xx(-4)t^(-5)+9xx(-2/3)t^(-5/3)`

= `48t^(-5)-6t^(-5/3)`

and accelaration at `t=1` is `48-6=42`

Jerk is `f'''(t)=48xx(-5)t^(-6)-6xx(-5/3)t^(-8/3)`

= `-240t^(-6)+10t^(-8/3)`

and jerk at `t=1` is `f'''(1)=-240+10=-230`

Answer 2

velocity at time t=1 sec is -3 `fps`
speed at t=1 sec is 3 `fps`
acceleration at time t = 1 sec is 42 `fp s^2`
jerk at time t=1 sec is -230 `fps^2`

Explanation:

Given,
`f(t)=4t^-3+27t^(1/3)+12`
To find
i) velocity
ii) speed
iii) acceleration
iv) jerk

velocity
`v=d/dtf(t)`
`f(t)=4t^-3+27t^(1/3)+12`
Differentiating f(t) wrt time t
`v=d/dt(4t^-3+27t^(1/3)+12)`
`d/dt(t^n)=nt^(n-1)`
For
`n=-3, n-1=-3-1=-4`
By constant rule
`d/dt(4t^-3)=4d/dt(t^-3)=(4)(-3)t^(-4)=-12t^-4`
For
`n=1/3, n-1=1/3-1=-2/3`
`d/dt(27t^(1/3))=27d/dt(t^(1/3))=(27)(1/3)t^(-2/3)`
`d/dt(constant)=0`
`d/dt(12)=0`
By sum rule,
`d/dt(4t^-3+27t^(1/3)+12)=d/dt(4t^-3)+d/dt(27t^(1/3))+d/dt(12)`
Substituting the values
`v=-12t^-4+(27)(1/3)t^(-2/3)+0`
Simplifying,
`v=-12t^-4+9t^(-2/3)`
At t=1,
`v=-12(1)^-4+9(1)^(-2/3)`
`v=-12+9`
`v=-3`
Thus, the velocity at time `t=1 sec, is -3 fps`

speed
speed = abs(velocity)
velocity at t=1 sec is -3 fps
abs(velocity)=abs(-3fps)
Thus, speed at t=1 sec is 3 fps

acceleration
a=`d/dtv`
`v=-12t^-4+9t^(-2/3)`
Differentiating v with respect to time t,
`d/dtv=d/dt(-12t^-4+9t^(-2/3))`
`d/dt(t^n)=nt^(n-1)`
For
`n=-4, n-1=-4-1=-5`
By constant rule
`d/dt(-12t^-4)=-12d/dt(t^-4)=(-12)(-4)t^(-5)=48t^-5`
For
`n=-2/3, n-1=-2/3-1=-5/3`
`d/dt(9t^(-2/3))=9d/dt(t^(-2/3))=(9)(-2/3)t^(-5/3)=-6t^(-5/3)`
By sum rule,
`d/dt(-12t^-4+9t^(-2/3))`=`d/dt(-12t^-4)+d/dt(9t^(-2/3))`
`=48t^-5-6t^(-5/3)`
Simplifying,
`a=48t^-5-6t^(-5/3)`
At time t = 1,
`a=48(1)^-5-6(1)t^(-5/3)`
`=48-6`
Thus, the acceleration at time`t = 1 sec`, is 42fp `s^2`

jerk

b=`d/dta`
Differentiating a wrt t
`d/dta=d/dt(48t^-5-6t^(-5/3))`
`d/dt(t^n)=nt^(n-1)`
For
`n=-5, n-1=-5-1=-6`
By constant rule
`d/dt(48t^-5)=48d/dt(t^-5)=(48)(-5)t^(-6)=-240t^-6`
For
`n=-5/3, n-1=-5/3-1=-8/3`
`d/dt(-6t^(-2/3))=-6d/dt(t^(-5/3))=(-6)(-5/3)t^(-8/3)=10t^(-8/3)`
By sum rule,
`d/dt(48t^-5-6t^(-5/3))=-240t^-6+10t^(-5/3)`
Simplifying,
`a=-240t^-6+10t^(-5/3)`
At time t=1, jerk is
`a=-240(1)^-6+10(1)^(-5/3)`
`a=-240+10`
`a=-230` `fps^2`