Respuesta :
Answer:
(1).The acceleration is -24 m/s²
The total distance is 912 m.
(2). The distance traveled before it stop is 78 m.
Explanation:
Given that,
Velocity = (12-3t^2) m/s
When t = 1 s, the particle is located 10 m to the left of the origin.
We need to calculate the acceleration at t = 4 sec
Using formula of acceleration
[tex]a=\dfrac{dv}{dt}[/tex]
Put the value of v
[tex]a=\dfrac{d}{dt}(12-3t^2)[/tex]
[tex]a=-6t[/tex]
Put the value of t
[tex]a=-6\times4[/tex]
[tex]a=-24\ m/s^2[/tex]
The displacement from t = 0,t = 10 s, and the distance the particle travels during this time period.
We need to calculate the distance
Using formula of distance
[tex]ds=v\ dt[/tex]
[tex]\int_{-10}^{s}=\int_{1}^{t}{v}dt[/tex]
Put the value of v
[tex]\int_{-10}^{s}=\int_{1}^{t}{ (12-3t^2)}dt[/tex]
[tex]s+10=12t-t^3-11[/tex]
[tex]s=12t-t^3-21[/tex]
At t = 0,
[tex]s_{t=0}=-21[/tex]
At t = 10,
[tex]s_{t=10}=12\times10-10^3-21[/tex]
[tex]s_{t=10}=-901[/tex]
The displacement is
[tex]\Delta s=-901-(-21)[/tex]
[tex]\Delta s=-880\ m[/tex]
The distance at t= 2 sec
[tex]s_{t=2}=12\times2-2^3-21[/tex]
[tex]s_{t=2}=-5[/tex]
The total distance will be,
[tex]s_{T}=(21-5)+(901-5)[/tex]
[tex]s_{T}=912\ m[/tex]
(2). We need to calculate the distance at 2 sec
Using equation of motion
[tex]s=ut-\dfrac{1}{2}at^2[/tex]
Put the value into the formula
[tex]s=27\times 2+\dfrac{1}{2}\times6\times(2)^3[/tex]
[tex]s=78\ m[/tex]
Hence, (1).The acceleration is -24 m/s²
The total distance is 912 m.
(2). The distance traveled before it stop is 78 m.
