Chapter 2: Q 17. (page 60)

A speed skater moving to the left across frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadily, then continues on at 6.0 m/s. What is her acceleration on the rough ice?

Short Answer

Expert verified

The acceleration of the skater is $- 2.8 m / s^{2}$

Step by step solution

Step 1. The skater is moving on the frictionless ice with speed 8.0 m/s Therefore, the initial speed of the skater is considered as vi→=8.0 m/s

The skater meets the rough patch of ice of width 5m.
Therefore, the distance travelled by skater on rough ice patch is $s = 5.0 m$
The speed of skater after crossing the ice patch is 6.0 m/s.
Hence, the final speed is considered as ${\vec{v}}_{f} = 6.0 m / s$ .
Let the acceleration of the skater is $a$ .

Step 2. The following figure represents the situation

Now, write the equation of motion,

${(\vec{v_{f}})}^{2} - {(\vec{v_{i}})}^{2} = 2 a s$
Substitute the values of all variables given to obtain acceleration $a$

role="math" localid="1648015090533" ${(6.0)}^{2} - {(8.0)}^{2} = 2 a (5.0) 36 - 64 = 10 a a = \frac{- 28}{10} a = - 2.8 m / s^{2}$

Therefore, the acceleration of the skater is $2.8 m / s^{2}$ . the minus sign represents that the speed of the skater has decreased and along with the speed, acceleration also decreases.