Question

A child on a sled starts from rest at the top of a 30.0° slope. If the trip to the bottom takes 15.0 s, how long is the slope? Assume that frictional forces may be neglected.

A) 293 m
B) 551 m
C) 795 m
D) 1130 m
E) 1470 m

Answer 1

The length of the slope that the child travels in 15.0 s is 551 meters, calculated using the kinematic equation for uniformly accelerated motion with a known angle and the acceleration due to gravity.

Step-by-step explanation:

To determine how long the slope is that a child on a sled will travel in 15.0 s, we need to calculate the distance using the kinematic equations for uniformly accelerated motion. Since friction is negligible and the sled starts from rest, we use the equation s = ut + (1/2)at², where s is the distance travelled, u is the initial speed (0 m/s in this case), a is the acceleration, and t is the time. The acceleration a can be found using the formula a = g · sin(θ), where g is the acceleration due to gravity (9.81 m/s²), and θ is the angle of the slope (30.0°). Substituting in the values and simplifying gives us:

a = 9.81 m/s² · sin(30.0°) = 9.81 m/s² · 0.5 = 4.905 m/s²

Now we can substitute a into the first equation to find the distance s:

s = (1/2) · 4.905 m/s² · (15.0 s)² = (1/2) · 4.905 m/s² · 225 s² = 551 m

Therefore, the length of the slope that the child travels is 551 meters, which corresponds to option B).