Question

practice

1. when a hockey player hits a hockey puck with his stick, the velocity of

the puck changes from 8.0 m/sn to 10.0 m/ss over a time interval
of 0.050 s. what is the average acceleration of the puck?
(overline { a } _ { mathrm { avg } } = 360 mathrm { m } / mathrm { s } ^ { 2 } mathrm { s } )

1. a racehorse takes 2.70 s to accelerate from a trot to a gallop. if the

horses initial velocity is 3.61 m/se and it experiences an acceleration
of ( 2.77 mathrm { m } / mathrm { s } ^ { 2 } mathrm { e } ), what is the racehorses final velocity when it gallops?
(overline { v } _ { f } = 11.1 mathrm { m } / mathrm { s } mathrm { e } )

Explanation:

Step1: Recall the formula for acceleration

The formula for average acceleration is \(a_{avg}=\frac{\Delta v}{\Delta t}\), where \(\Delta v = v_f - v_i\).

Step2: Assign directions and values

Let north be positive and south be negative. So \(v_i = 8.0\ m/s\) (positive as north - \(N\)), \(v_f=- 10.0\ m/s\) (negative as south - \(S\)), and \(\Delta t = 0.050\ s\).

Step3: Calculate \(\Delta v\)

\(\Delta v=v_f - v_i=-10.0 - 8.0=-18.0\ m/s\)

Step4: Calculate \(a_{avg}\)

Using \(a_{avg}=\frac{\Delta v}{\Delta t}\), substitute \(\Delta v=-18.0\ m/s\) and \(\Delta t = 0.050\ s\). So \(a_{avg}=\frac{-18.0}{0.050}=- 360\ m/s^{2}\). The negative sign indicates the direction is south.

For problem 4:

Step1: Recall the formula for final velocity

The formula for final velocity is \(v_f=v_i + a\Delta t\)

Step2: Substitute the values

Given \(v_i = 3.61\ m/s\) (east - \(E\), so positive), \(a = 2.77\ m/s^{2}\) (east - \(E\), positive) and \(\Delta t=2.70\ s\)
\(v_f=3.61+2.77\times2.70\)
First calculate \(2.77\times2.70 = 7.479\)
Then \(v_f=3.61 + 7.479=11.089\approx11.1\ m/s\) (east - \(E\))

Answer:

For problem 3: The average acceleration of the puck is \(360\ m/s^{2}\) south.
For problem 4: The racehorse’s final velocity is \(11.1\ m/s\) east.