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

Answer:

$D_T=18.567m$

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

 $V^2=2as$

 $V=\sqrt{2*6*1.05}$

 $V=4.1m/s$

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

 $d_3=\frac{u_3^2}{2ad_3}$

 $d_3=\frac{4.1^2}{2*1.64}$

 $d_3=5.125m$

Generally the Total Distance Traveled is mathematically given by

 $D_T=d_1+d_2+d_3$

 $D_T=5.125m+12.392+1.05 m$

 $D_T=18.567m$