Question

⦁ A speed skater increases her speed form 10 m/s to 12.5 m/s over a period of 3 seconds while coming out of a curve of 20 m radius. What are the magnitudes of her radial, tangential and total accelerations as she leaves the curve?

Answer 1

Answer:

7.85 m/s^2

Explanation:

linear or tangential acceleration= dv/dt

⇒$a_t= \frac{12.5-10}{3}$

=0.83 m/s^2

radial acceleration is given by = $\frac{v^2}{r}$

⇒$a_r =\frac{12.5^2}{20}$

= 7.81 m/s^2

total acceleration

$a_T= \sqrt{a_t^2+a_r^2}$

putting values we get

$a_T= \sqrt{0.83^2+7.81^2}$

= 7.85 m/s^2

Answer 2

Answer:

Explanation:

Tangential acceleration = ( 12.5 - 10 )/ 3

a_t= .833 m /s²

radial acceleration

= v² / R

12.5² / 20 ( 12.5 m/s is the velocity when it leaves the curve )

a_r= 7.81 ms⁻²  

Total acceleration √( .833² + 7.81²)

=  √( 61.6939)

= 7.85 m/s⁻