Question

An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object then slides at a constant velocity for 6.0 s until it reaches a rough section that causes the object to stop in 2.5 s.

Answer 1

`D_T=18.567m`

Step-by-step explanation:

From the question we are told that:

Acceleration
`a=8.0 m/s^2`

Displacement
`d=1.05 m`

Initial time
`t_1=6.0s`

Final Time
`t_2=2.5s`

Generally the equation for Velocity of 1.05 travel is mathematically given by

Using Newton's Law of Motion

`V^2=2as`

`V=√(2*6*1.05)`

`V=4.1m/s`

Generally the equation for Distance traveled before stop is mathematically given by

`d_2=v*t_1`

`d_2=3.098*4`

`d_2=12.392`

Generally the equation for Distance to stop is mathematically given by

Since For this Final section

Final velocity
`v_3=0 m/s`

Initial velocity
`u_3=4.1 m/s`

Therefore

Using Newton's Law of Motion

`-a_3=(4.1)/(2.5)`

`-a_3=1.64m/s^2`

Giving

`v_3^2=u^2-2ad_3`

Therefore

`d_3=(u_3^2)/(2ad_3)`

`d_3=(4.1^2)/(2*1.64)`

`d_3=5.125m`

Generally the Total Distance Traveled is mathematically given by

`D_T=d_1+d_2+d_3`

`D_T=5.125m+12.392+1.05 m`

`D_T=18.567m`