Given:
The sled decelerates with a uniform rate of: a = -0.43 m/s^2 (The negative sign indicates that the velocity of the sled is decreasing)
The sled covers a distance of 85 m before coming to rest.
To find:
The time sled takes to come to rest
Step-by-step explanation:
We make use of the following kinematic equation,

Here, v is the final velocity, u is the initial velocity, a is the deceleration and s is the displacement.
The final velocity v is zero as the sled is at rest. Thus, v = 0 m/s
The displacement of the sled after hitting the breaks is: s = 85 m
Substituting the values in the above equation, we get:

The initial velocity of the sled when breaks are applied is 8.55 m/s.
Now, consider another kinematical equation,

Substituting the values in the above equation, we get:

Final answer:
The sled takes 19.88 seconds before it comes to rest.