Can someone please tell me if I've approached this homework problem correctly. A boy is running at his best speed of 6.0 m/s to catch a bus stopped at the lights. When he is 16 m from the bus, the light goes green and the bus pulls away at an acceleration of 1.0 m/s/s. Does he catch up with the bus and if yes how far does he have to run?
I set the distance of the boy equal to the distance of the bus in order for him to catch up to it and got: 6t=16 +1/2t^2, then solved the quad. equation to get t=4 or 8. So he ran 24m till he caught the bus. Did I do this correctly.
That is the right equation. I will rewrite it as:
t^2 -12t + 32 = 0
(t-4)(t-8) = 0
(The t=8 time is when the bus would pull past him, if he didn't get on and passed it at t=4)
He will have run 24 s at t=4. Yes, you did it right. Nice job.
Yeah! thank you very much for checking my work. I can go to bed now.
I meant 24 meters, not 24 seconds. You figured that out already
Yes, you have approached the homework problem correctly. To determine whether the boy catches up with the bus, you correctly set the distance traveled by the boy equal to the distance traveled by the bus.
Let's break down your solution step by step:
1. You start by equating the distance traveled by the boy with the distance traveled by the bus. The distance traveled by the boy is given by the equation: 6t, where t is the time in seconds. The distance traveled by the bus is given by the equation: 16 + (1/2)at^2, where a is the acceleration of the bus.
2. By setting these two distances equal, you obtain the equation: 6t = 16 + (1/2)at^2.
3. The next step is to solve the quadratic equation to find the possible values of t. Rearranging the equation, you get: (1/2)at^2 - 6t + 16 = 0.
4. You can then apply the quadratic formula to solve for t: t = (-b ± √(b^2 - 4ac)) / (2a), where a = (1/2)a, b = -6, and c = 16.
5. Plugging in the values, you get: t = (-(-6) ± √((-6)^2 - 4*(1/2)*16)) / (2*(1/2)*a). Simplifying further, you get: t = (6 ± √(36 - 32a)) / a.
6. Since we know the acceleration of the bus is 1.0 m/s^2, you substitute a = 1.0 into the equation: t = (6 ± √(36 - 32*1.0)) / 1.0. Simplifying, you get: t = (6 ± √4) / 1.0.
7. This results in two possible values for t: t = (6 + 2) / 1.0 = 8 seconds, and t = (6 - 2) / 1.0 = 4 seconds.
8. Therefore, the boy has two possible times at which he catches up with the bus, either 4 seconds or 8 seconds.
9. Lastly, you calculated the distance traveled by the boy in each case. When t = 4 seconds, the distance traveled by the boy is 6 * 4 = 24 meters. Similarly, when t = 8 seconds, the distance traveled by the boy is 6 * 8 = 48 meters.
Therefore, he runs a total of 24 meters until he catches the bus, which is the correct answer. Well done!