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

Step-by-step 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⁻

Answer 2

7.85 m/s^2

Step-by-step explanation:

linear or tangential acceleration= dv/dt

⇒
`a_t= (12.5-10)/(3)`

=0.83 m/s^2

radial acceleration is given by =
`(v^2)/(r)`

⇒
`a_r =(12.5^2)/(20)`

= 7.81 m/s^2

total acceleration

`a_T= √(a_t^2+a_r^2)`

putting values we get

`a_T= √(0.83^2+7.81^2)`

= 7.85 m/s^2