Question

A hockey player skating at a velocity of 15.0 m/s starts sliding and slows at a rate of – 0.42 m/s 2 . During the first 5 seconds, how far has he traveled? How long does he take to stop? What is his velocity at t=8.0 s?

Answer 1

Answer:

Explanation:

Given:

• Initial velocity of player, u = 15.0 m/s
• Acceleration of player, a = -0.42 m/s²

To find

• Distance covered by player in first t = 5 sec; s =?
• Time taken by boy to come to rest; T =?
• Velocity of boy at t = 8.0 sec ; v =?

Knowledge required

• First equation of motion

     v = u + a t

• Second equation of motion

     s = u t + 1/2 a t²

[ Where v represent final velocity, u represent initial velocity, s is distance covered, t is time taken, a represent acceleration ]

Solution

• Calculating distance covered by player in first 5 sec (s =?)

Using second equation of motion

→ s = u t + 1/2 a t²

→ s = ( 15 ) ( 5 ) + 1/2 ( -0.42 ) ( 5 )²

→ s = 75 - 5.25

→ s = 69.75 m

• Calculating time taken by player to stop (T=?)

Using first equation of motion

→ v = u + a T

→ 0 = 15 + (-0.42 ) T

→ -15 = -0.42 T

→ T = 15/0.42

→ T = 35.71 sec

• Calculating velocity of player at t =8.0 sec (v=?)

Using first equation of motion

→ v = u + a t

→ v = (15) + ( -0.42 ) ( 8 )

→ v = 15 + 3.36

→ v = 18.36  m/s

Therefore,

• Player will travel 69.75 metres in first 5 seconds.

• Player took 35.71 seconds approximately to come to rest.

• Velocity of player at t = 8 s would be 18.36 m/s.