A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The to
tal mass of the sled, man, and rock is 96.0 kg. The mass of the rock is 0.310 kg and the man can throw it with a speed of 14.5 m/s. Both speeds are relative to the ground. Determine the speed of the sled if the man throws the rock forward (i.e. in the direction the sled is moving). Also determine the speed of the sled if he throws the rock directly backwards.
Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:
Since the car starts from rest, v₀ =0.
We know the value of t = 5 sec., but we need to find the value of a.
Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:
Replacing a and t in (1):
Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:
Replacing v₀, at and t in (1), we have:
Therefore, as the truck travels twice as far as the car, the right answer is a).
