Answer to Question #85768 in Mechanics | Relativity for Charlie

Question

An object with an initial speed of 4.21 m/s accelerated at a speed of 2.45 m/s^2 over a distance of 12.3 m across a sheet of ice. The object then hits a rough patch causeing the object to come to rest at 5.82 s. How long was the object accelerating for on the ice? What was the rate of acceleration of the object when sliding on the rough section? What was the total displacement of the object throughout the entire trip?

Answer 1

The distance

$$d_1=v_i t+\frac{at^2}{2}\\ 12.3=4.21t+\frac{2.45t^2}{2}$$

So, the time of the object motion for on the ice

$$t=1.89\;\rm{s}$$

The final velocity on the ice

$$v_f=v_i+at=4.21+2.45\times 1.89=8.84\;\rm{m/s}$$

The rate of acceleration of the object when sliding on the rough section

$$a=\frac{-8.84}{5.82}=-1.52\;\rm{m/s^2}$$

The total displacement of the object throughout the entire trip

$$d=d_1+d_2=12.3+\frac{8.84+0}{2}\times 5.82=38\;\rm{m}$$