A skateboarder is performing tricks on a specially designed ramp. The height, [tex]h[/tex], in meters, of the skateboard above the ground can be modeled by the function [tex]h(t)=2 t^2-8 t+10[/tex], where [tex]t[/tex] represents the time in seconds since the skateboarder started the trick. Remember to show your work for full credit.
a. Find the vertex of the quadratic function and interpret its meaning in the context of the problem.
b. Determine the height of the skateboard when the skateboarder starts the trick. Explain.
c. Determine when the skateboard will be at ground level.
d. Determine the range of times during which the skateboard will be at least 5 meters above the ground.
e. What would be the height of the skateboard after 3 seconds?
Answer (1)
Finds the vertex of the quadratic function: t v = 2 , h ( 2 ) = 2 .
Determines the initial height: h ( 0 ) = 10 .
Concludes the skateboard never reaches ground level due to a negative discriminant.
Calculates the time intervals when the skateboard is at least 5 meters above ground: t ≤ 0.775 and t ≥ 3.225 .
Finds the height at 3 seconds: h ( 3 ) = 4 .
h ( 3 ) = 4
Explanation
Problem Analysis We are given the height function h ( t ) = 2 t 2 − 8 t + 10 which models the skateboarder's height above the ground at time t . We need to analyze this function to answer several questions about the skateboarder's motion.
Finding the Vertex To find the vertex of the quadratic function h ( t ) = 2 t 2 − 8 t + 10 , we first find the time t at which the vertex occurs using the formula t v = − b / ( 2 a ) , where a = 2 and b = − 8 . Thus, t v = − ( − 8 ) / ( 2 ∗ 2 ) = 8/4 = 2 . Now, we find the height at this time by plugging t v = 2 into the height function: h ( 2 ) = 2 ( 2 ) 2 − 8 ( 2 ) + 10 = 2 ( 4 ) − 16 + 10 = 8 − 16 + 10 = 2 . Therefore, the vertex is at ( 2 , 2 ) . In the context of the problem, this means that the minimum height of the skateboarder is 2 meters, and it occurs at 2 seconds.
Initial Height To determine the height of the skateboard when the skateboarder starts the trick, we need to find the height at t = 0 . Plugging t = 0 into the height function, we get: h ( 0 ) = 2 ( 0 ) 2 − 8 ( 0 ) + 10 = 0 − 0 + 10 = 10 . So, the height of the skateboarder when they start the trick is 10 meters.
Time at Ground Level To determine when the skateboard will be at ground level, we need to find the time t when h ( t ) = 0 . This means solving the quadratic equation 2 t 2 − 8 t + 10 = 0 . We can use the quadratic formula: t = 2 a − b ± b 2 − 4 a c , where a = 2 , b = − 8 , and c = 10 . Plugging in these values, we get: t = 2 ( 2 ) 8 ± ( − 8 ) 2 − 4 ( 2 ) ( 10 ) = 4 8 ± 64 − 80 = 4 8 ± − 16 . Since the discriminant is negative, there are no real solutions for t . This means the skateboard never reaches ground level.
Time Above 5 Meters To determine the range of times during which the skateboard will be at least 5 meters above the ground, we need to solve the inequality h ( t ) ≥ 5 . This means 2 t 2 − 8 t + 10 ≥ 5 , which simplifies to 2 t 2 − 8 t + 5 ≥ 0 . First, we find the roots of the quadratic equation 2 t 2 − 8 t + 5 = 0 using the quadratic formula: t = 2 a − b ± b 2 − 4 a c = 2 ( 2 ) 8 ± ( − 8 ) 2 − 4 ( 2 ) ( 5 ) = 4 8 ± 64 − 40 = 4 8 ± 24 = 4 8 ± 2 6 = 2 ± 2 6 . So the roots are t 1 = 2 − 2 6 and t 2 = 2 + 2 6 . Since the parabola opens upwards, the skateboard will be at least 5 meters above the ground for times t ≤ 2 − 2 6 and t ≥ 2 + 2 6 . Approximating the values, we have t 1 ≈ 2 − 2 2.45 ≈ 2 − 1.225 ≈ 0.775 and t 2 ≈ 2 + 1.225 ≈ 3.225 . Therefore, the skateboard is at least 5 meters above the ground for t ≤ 0.775 seconds and t ≥ 3.225 seconds.
Height After 3 Seconds To find the height of the skateboard after 3 seconds, we plug t = 3 into the height function: h ( 3 ) = 2 ( 3 ) 2 − 8 ( 3 ) + 10 = 2 ( 9 ) − 24 + 10 = 18 − 24 + 10 = 4 . So, the height of the skateboard after 3 seconds is 4 meters.
Final Answer In summary: a. The vertex of the quadratic function is ( 2 , 2 ) , which means the minimum height of the skateboarder is 2 meters at 2 seconds. b. The height of the skateboard when the skateboarder starts the trick is 10 meters. c. The skateboard never reaches ground level. d. The skateboard is at least 5 meters above the ground for t ≤ 0.775 seconds and t ≥ 3.225 seconds. e. The height of the skateboard after 3 seconds is 4 meters.
Examples
Understanding quadratic functions like the one in this problem can help in various real-world scenarios. For instance, designing a roller coaster involves analyzing parabolic paths to ensure safety and excitement. Similarly, in sports, understanding the trajectory of a ball or projectile often involves quadratic equations. By modeling the height and distance of objects using these functions, engineers and athletes can optimize performance and design safer, more efficient systems. This problem showcases how mathematical models can provide valuable insights into real-world phenomena, enabling better decision-making and innovation.
