Question

A particle travels along a straight line with a velocity v = (12 - 3t) m/s, where t is in seconds. When t = 1 s, the particle is located 10 m to the left of the origin. Determine the acceleration when t = 4 s. the displacement from t = 0,t = 10 s, and the distance the particle travels during this time period. a 9 2. A sphere is fired downwards into a medium with an initial speed of 27 m/s. If it experiences a deceleration of a = (-6t) m/s2 ,where t is in seconds, determine the distance traveled before it stops.

Answer 1

(1).The acceleration is -24 m/s²

The total distance is 912 m.

(2). The distance traveled before it stop is 78 m.

Step-by-step 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

`a=(dv)/(dt)`

Put the value of v

`a=(d)/(dt)(12-3t^2)`

`a=-6t`

Put the value of t

`a=-6*4`

`a=-24\ m/s^2`

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

`ds=v\ dt`

`\int_(-10)^(s)=\int_(1)^(t){v}dt`

Put the value of v

`\int_(-10)^(s)=\int_(1)^(t){ (12-3t^2)}dt`

`s+10=12t-t^3-11`

`s=12t-t^3-21`

At t = 0,

`s_(t=0)=-21`

At t = 10,

`s_(t=10)=12*10-10^3-21`

`s_(t=10)=-901`

The displacement is

`\Delta s=-901-(-21)`

`\Delta s=-880\ m`

The distance at t= 2 sec

`s_(t=2)=12*2-2^3-21`

`s_(t=2)=-5`

The total distance will be,

`s_(T)=(21-5)+(901-5)`

`s_(T)=912\ m`

(2). We need to calculate the distance at 2 sec

Using equation of motion

`s=ut-(1)/(2)at^2`

Put the value into the formula

`s=27* 2+(1)/(2)*6*(2)^3`

`s=78\ m`

Hence, (1).The acceleration is -24 m/s²

The total distance is 912 m.

(2). The distance traveled before it stop is 78 m.