Question

A cross-country skier is skiing along at a zippy 8.0 m/s. She stops pushing and simply glides along, slowing to a reduced speed of 6.0 m/s after gliding for 5.0 m. What is the magnitude of her acceleration as she slows

Answer 1

Answer:

 The magnitude of acceleration cross-country skier as she slows = $1.2 m/s^2$

Explanation:

We have equation of motion, $v^2=u^2+2as$, where u is the initial velocity, u is the final velocity, s is the displacement and a is the acceleration.

In this case we have final velocity = 6.0 m/s, initial velocity = 8.0 m/s, and displacement = 5.0 m, now we need to find acceleration value.

Substituting

  $6^2=8^2+2*a*5\\ \\ a=-1.2 m/s^2$

  We have acceleration value = $-1.2 m/s^2$

  Magnitude of acceleration =$1.2 m/s^2$

 The magnitude of acceleration cross-country skier as she slows = $1.2 m/s^2$

Answer 2

Answer:

The magnitude of her acceleration is -2.8 m/s²

Explanation:

It is given that,

Initial velocity of the skier, u = 8 m/s

Final velocity of the skier, v = 6 m/s

Distance covered, s = 5 m

We need to find the magnitude of acceleration as she slows down. It can be calculated using third equation of motion as :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(6\ m/s)^2-(8\ m/s)^2}{2\times 5\ m}$

$a=-2.8\ m/s^2$

Answer:

The magnitude of her acceleration is -2.8 m/s²

Explanation:

It is given that,

Initial velocity of the skier, u = 8 m/s

Final velocity of the skier, v = 6 m/s

Distance covered, s = 5 m

We need to find the magnitude of acceleration as she slows down. It can be calculated using third equation of motion as :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(6\ m/s)^2-(8\ m/s)^2}{2\times 5\ m}$

$a=-2.8\ m/s^2$

So, the acceleration of the skier is -2.8 m/s². Hence, this is the required solution.