A ball bearing is placed on an inclined plane and begins to roll. The angle of elevation of the plane is [tex]$\theta$[/tex]. The distance (in meters) the ball bearing rolls in [tex]$t$[/tex] seconds is [tex]$s(t)=4.9(\sin (\theta)) t^2$[/tex].
(a) Determine the speed (in meters per second) of the ball bearing after [tex]$t$[/tex] seconds.
(b) Complete the table.
| | | [tex]$\frac{\pi}{4}$[/tex] | [tex]$\frac{\pi}{3}$[/tex] | [tex]$\frac{\pi}{2}$[/tex] |
| :-----------------: | :-: | :-----------------: | :-----------------: | :-----------------: |
| [tex]$s^{\prime}(t)$[/tex] | | | | |
| [tex]$\theta$[/tex] | [tex]$\frac{2 \pi}{3}$[/tex] | [tex]$\frac{3 \pi}{4}$[/tex] | [tex]$\pi$[/tex] |
| :-----------------: | :-----------------: | :-----------------: | :-----------------: |
| [tex]$s^{\prime}(t)$[/tex] | | | |
Use the table to determine the value of [tex]$\theta$[/tex] that produces the maximum speed at a particular time.
Answer (1)
Find the speed by taking the derivative of the distance function: s ′ ( t ) = 9.8 t sin ( θ ) .
Calculate the speed for different values of θ : 0 , 4 π , 3 π , 2 π , 3 2 π , 4 3 π , π .
Determine that the maximum speed occurs when sin ( θ ) is maximized, which is when θ = 2 π .
Conclude that the angle that produces the maximum speed is 2 π .
Explanation
Problem Setup The distance the ball bearing rolls is given by the function s ( t ) = 4.9 ( sin ( θ )) t 2 , where θ is the angle of elevation and t is the time in seconds.
Finding the Speed To find the speed, we need to find the derivative of s ( t ) with respect to t . Using the power rule, we have:
s ′ ( t ) = d t d [ 4.9 ( sin ( θ )) t 2 ] = 4.9 ( sin ( θ )) ⋅ 2 t = 9.8 t sin ( θ )
So, the speed of the ball bearing after t seconds is 9.8 t sin ( θ ) m/s.
Completing the Table Now, we need to complete the table for s ′ ( t ) for different values of θ . We'll calculate s ′ ( t ) for θ = 0 , 4 π , 3 π , 2 π , 3 2 π , 4 3 π , π . We will assume t = 1 for simplicity, as the question asks for the value of θ that produces the maximum speed at a particular time. The value of t does not affect the angle at which the speed is maximized.
Calculating Speeds for Different Angles For θ = 0 , s ′ ( t ) = 9.8 t sin ( 0 ) = 0 .
For θ = 4 π , s ′ ( t ) = 9.8 t sin ( 4 π ) = 9.8 t ⋅ 2 2 ≈ 6.9296 t .
For θ = 3 π , s ′ ( t ) = 9.8 t sin ( 3 π ) = 9.8 t ⋅ 2 3 ≈ 8.4870 t .
For θ = 2 π , s ′ ( t ) = 9.8 t sin ( 2 π ) = 9.8 t ⋅ 1 = 9.8 t .
For θ = 3 2 π , s ′ ( t ) = 9.8 t sin ( 3 2 π ) = 9.8 t ⋅ 2 3 ≈ 8.4870 t .
For θ = 4 3 π , s ′ ( t ) = 9.8 t sin ( 4 3 π ) = 9.8 t ⋅ 2 2 ≈ 6.9296 t .
For θ = π , s ′ ( t ) = 9.8 t sin ( π ) = 9.8 t ⋅ 0 = 0 .
Finding the Angle for Maximum Speed To find the value of θ that produces the maximum speed at a particular time t , we need to maximize sin ( θ ) . The maximum value of sin ( θ ) is 1, which occurs when θ = 2 π . Therefore, the maximum speed is achieved when θ = 2 π .
Final Answer The speed of the ball bearing after t seconds is 9.8 t sin ( θ ) m/s. The completed table (assuming t = 1 ) is:
θ
0
4 π
3 π
2 π
3 2 π
4 3 π
π
s ′ ( t )
0
4.9 2
4.9 3
9.8
4.9 3
4.9 2
0
The value of θ that produces the maximum speed at a particular time is 2 π .
Conclusion The speed of the ball bearing is given by s ′ ( t ) = 9.8 t sin ( θ ) . The angle that maximizes the speed is θ = 2 π .
Examples
Imagine you're designing a ramp for a skateboarding competition. You need to determine the angle of the ramp that will allow the skateboarders to reach the maximum speed at a certain point. By using the formula s ( t ) = 4.9 ( sin ( θ )) t 2 and finding the derivative to get the speed s ′ ( t ) = 9.8 t sin ( θ ) , you can calculate the optimal angle θ to maximize the skateboarder's speed and ensure an exciting competition. This is a practical application of calculus in sports engineering and design.
