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Acceleration, velocity, time question in basic physics? |
A bicyclist is finishing his repair of a flat tire when a friend rides by at 3.5 m/s. Two seconds later, the bicyclist hops on his bike and accelerates at 3.0 m/s2 until he catches his friend. How much time (in seconds) does it take until he catches his friend? First create two equations which describe the motion of the bicyclers. The first bicycler rides at a constant speed (3.5 m/s). The distance traveled by this bicycler is given as, d = v * t where d is the distance traveled, v is his constant speed, and t is the time traveled. The second bicycler starts from rest but maintains a constant acceleration. The 2nd bicycler's distance traveled is given as, d = 1/2 a * t^2 where d is, again, the distance traveled, a is the constant acceleration (3.0 m/s^2), and t is the time the bicycler is riding. When the 2nd bicycler catches up with his friend, both of their distances will be the same (d for the first rider = d for the second rider), so we can set their equations equal to each other with one slight change. The time, t, in each equation is not the same time t. The first rider had a two second head start on the second rider. In this time of 2 seconds, the rider moved a distance of 7 meters (= 3.5 m/s * 2 s) away before the 2nd rider even started to move. So to compensate for this fact we will add on 7 meters (as an initial distance away) to the first rider鈥檚 equation so that the two times, t, will match up. So we set the equations equal to each other (with the slight modification), 3.5 m/s * t + 7 meters = 1/2 * 3.0 m/s^2 * t^2 Where t is the time after the second rider starts moving (2 seconds after the first rider passes). We now merely solve for the time t, which will require the use of the quadratic equation. ax^2 + bx + c = 0 x = (-b +/- sqrt (b^2 鈥?4ac)) / 2a When we do this, we get two values for time鈥 positive and a negative value. We are only concerned with the positive value right now. Solving the equation gives us a time of 3.62 second. That means that 3.62 seconds after the 2nd rider starts to move he will catch his friend (5.62 seconds past when the friend rides by). We can check this answer by plugging this value back into the above equations and making sure they are equal. 3.5 m/s * (3.62 s) + 7 meters = 19.67 meters 1/2 * 3.0 m/s^2 * (3.62 s)^2 = 19.66 meters鈥hich is essentially equal (it would be exactly equal if I had carried over more significant figures from the time). Calculating how far the friend has gone after 2 seconds: Vfriend = 3.5 m/sec 鈭員 = 2 sec V = (鈭咲/鈭員) 3.5 m/sec = (鈭咲) / (2 sec) 鈭咲 = 7 m, how far the friend went after 2 seconds. He still has a constant velocity of 3.5 m/sec. aB (acceleration of bicyclist) = 2.8 m/sec虏 VoB = 0 鈭咲 = Vo鈭員 + (1/2)a鈭員虏 For the accelerating bycyclist: 鈭咲B = (0.5) aB 鈭員虏 鈭咲B = (0.5) (2.8 m/sec虏) 鈭員虏 鈭咲B = (1.4 m/sec虏) 鈭員虏 Now let's look at the friend's displacement: 鈭咲 = Vo鈭員 + (1/2)a鈭員虏 鈭咲F = V鈭員 鈭咲F = (3.5 m/sec) 鈭員 We know this from before: 鈭咲B = 鈭咲F + 7m (F = friend) Substitute 鈭咲F and 鈭咲B into the equation: (1.4 m/sec虏) 鈭員虏 - 7m = (3.5 m/sec) 鈭員 1.4 m/sec虏 (鈭員虏) - 3.5 m/sec (鈭員) - 7m = 0 Using the quadratic formula, I get: 鈭員 = 3.8117 sec OR -1.3117 sec Obviously, it can't be negitive, so we are left with the answer: 鈭員 = 3.8117 sec |
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