Practice Problems -continued 21. A professional baseball pitcher pitches a ball giving it a velocity of 45m/s toward the batter The batted ball ha a velocity of 30m/s toward the pitcher. Let the directio from the batter to the pitcher be the positive direction. change in velocity takes place over a period of 1.2 s ,wh the average acceleration of the baseball? 2. A child rolled a ball up a hill . At time zero, the ball had velocity of 1.8m/s up the hill. After 6.5 s the ball's vel was 2.3m/s down the hill . Let uphill be the positive direction. What was the average acceleration of the ball is the meaning of the sign of the acceleration? 3. Objects near Earth's surface fall with an acceleration of 9.81m/s^2 . If you dropped a rock from a cliff over a ri how fast would the rock be falling 4.I s after you dro A. The average acceleration of the space shuttle at takeof 29m/s^2[up] . What is the shuttle's velocity after 12 s? Let up be the positive direction. 5. A car is initially travelling at a velocity of 4.2m/s[W car's average acceleration is 0.86m/s^2[W] how long take the car to reach a velocity of 9.6m/s[W]

**21.** The initial velocity is *v₀* = -45 m/s (toward the batter, so negative). The final velocity is *v* = 30 m/s. The time interval is *Δt* = 1.2 s. Average acceleration is given by:<br /><br />*a* = ( *v* - *v₀* ) / *Δt*<br />*a* = (30 m/s - (-45 m/s)) / 1.2 s<br />*a* = (75 m/s) / 1.2 s<br />*a* = 62.5 m/s²<br /><br />**22.** Initial velocity *v₀* = 1.8 m/s (uphill, positive). Final velocity *v* = -2.3 m/s (downhill, negative). Time interval *Δt* = 6.5 s.<br /><br />*a* = (*v* - *v₀*) / *Δt*<br />*a* = (-2.3 m/s - 1.8 m/s) / 6.5 s<br />*a* = (-4.1 m/s) / 6.5 s<br />*a* ≈ -0.63 m/s²<br /><br />The negative sign indicates the average acceleration is downhill. The ball is slowing down while going uphill and then speeds up going downhill.<br /><br />**23.** Initial velocity *v₀* = 0 m/s (dropped from rest). Acceleration *a* = 9.81 m/s². Time *t* = 4.1 s. We can use the following equation:<br /><br />*v* = *v₀* + *at*<br />*v* = 0 m/s + (9.81 m/s²)(4.1 s)<br />*v* ≈ 40.2 m/s downward<br /><br />**24.** Initial velocity *v₀* = 0 m/s (starts from rest). Acceleration *a* = 29 m/s² [up]. Time *t* = 12 s.<br /><br />*v* = *v₀* + *at*<br />*v* = 0 m/s + (29 m/s²)(12 s)<br />*v* = 348 m/s [up]<br /><br />**25.** Initial velocity *v₀* = 4.2 m/s [W]. Final velocity *v* = 9.6 m/s [W]. Acceleration *a* = 0.86 m/s² [W]. We are looking for time *t*.<br /><br />*v* = *v₀* + *at*<br />*t* = (*v* - *v₀*) / *a*<br />*t* = (9.6 m/s - 4.2 m/s) / 0.86 m/s²<br />*t* = (5.4 m/s) / 0.86 m/s²<br />*t* ≈ 6.3 s<br />