Question

A loaded 350 kg toboggan is traveling on smooth horizontal snow at 4.30 m/s when it suddenly comes to a rough region. The region is 6.20 m long and reduces the toboggan's speed by 1.40 m/s. What is the average acceleration of the toboggan during this time?

1) 0.23 m/s²
2) 0.24 m/s²
3) 0.25 m/s²
4) 0.26 m/s²

Answer 1

The average acceleration of the toboggan through the rough region can be calculated using the kinematic equation, and is found to be -0.8129 m/s², indicating deceleration.

Step-by-step explanation:

To find the average acceleration of the toboggan as it moves through the rough region, we can use the kinematic equation related to velocity, acceleration, and distance: v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

The initial velocity u is 4.30 m/s, and the final velocity v after passing through the rough region is 4.30 m/s - 1.40 m/s = 2.90 m/s. The distance s is given as 6.20 m. Plugging these values into the equation and solving for a gives:

v² = u² + 2as
2.90² = 4.30² + 2 * a * 6.20
8.41 = 18.49 - 12.4a
12.4a = 18.49 - 8.41
12.4a = 10.08
a = 10.08 / 12.4
a = 0.8129 m/s²

However, since the toboggan is slowing down, the acceleration is negative. Thus, the average acceleration is -0.8129 m/s².