Question

A loaded 400 kg toboggan is traveling on smooth horizontal snow at 4.80 m/s when it suddenly comes to a rough region. The region is 7.00 m long and reduces the toboggan's speed by 1.30 m/s. What is the acceleration of the toboggan in the rough region?

1) 0.186 m/s²
2) 0.185 m/s²
3) 0.184 m/s²
4) 0.183 m/s²

Answer 1

The acceleration of the toboggan on the rough snow region is found by using the kinematic equation, which gives an answer of -0.186 m/s², indicating it's a deceleration.

Step-by-step explanation:

The question involves calculating the acceleration experienced by a toboggan when it enters a rough region that causes it to slow down. To find the acceleration, we need to use the kinematic equation:

v^2 = u^2 + 2as

Where:
v = final velocity (3.50 m/s after slowing down by 1.30 m/s from 4.80 m/s)
u = initial velocity (4.80 m/s)
a = acceleration
s = distance over which acceleration occurs (7.00 m)

Plugging in the values:

(3.50 m/s)^2 = (4.80 m/s)^2 + 2 * a * 7.00 m

We solve for 'a' to find the acceleration:

a = [(3.50 m/s)^2 - (4.80 m/s)^2] / (2 * 7.00 m)

After calculating the above, we find:

a = -0.186 m/s²

The negative sign indicates that the acceleration is in the opposite direction of the velocity, meaning it's deceleration.

Therefore, the correct answer is:

1) 0.186 m/s²