Ice Hockey Sticks Rated
The acceleration is equal to 0 if no acceleration in this problem?
A hockey player has a slapshot on a disk based on the ice. Suppose: The contact time is 0.0133 s. The disc blades on the ice of 12.2 feet without friction, when running on a surface concrete. Disk, and then decelerates at a uniform rate of 10.5 ft/s2. If the disk speed is 24.9 m / s after traveling 126 meters from the impact point, which is the average acceleration provided on the disc by the hockey stick? ft/s2 response units. There no friction, what then there is no acceleration? or am I look so bad? Ok, so the acceleration then the disk is slowing down. I must find a middle initial acceleration and deceleration (acceleration neg.) to get the average. acceleration is requested? Oh, and I asked if there is no friction effect of the acceleration. The acceleration is equal to 0 if there is no friction in this problem? Oops
It is a difficult problem because there are three distinct phases in the movement the disc, and it must be clear that this is speaking. In my opinion, "the average acceleration provided by the hockey stick" is refers only to the acceleration really suffered any contact the club (ie, during the first 0.0133 seconds). Step 1: Stick is in contact with the disc. In this phase ———————————————– : Publication rate: final Vi_1 = 0 Speed: Vf_1 =? Distance traveled: d1 =? Time Elapsed: 0.0133 acceleration a1 = t1 =? Notes on Phase 1: By definition acceleration, a1 = (Vf_1 – Vi_1) / T1. But we can not solve yet because we do not know Vf_1. Phase 2: Puck slides on the ice. In this phase ———————————————– : Starting rate: Vi_2 =? Final speed: Vf_2 =? Distance: 12.2 m d2 = Time elapsed: t2 =? Acceleration: a2 = 0 Notes on Phase 2: * We know that a2 = 0 because there is no friction this phase. No friction means no power means no acceleration. * A2 = 0 means (Vf_2 – Vi_2) / T2 = 0. And that means Vf_2 = Vi_2 (ie, speed does not change at this point. * At this point, the disc begins with the same speed he had at the end of Phase 1. This means Vi_2 = Vf_1. * Therefore, Vf_1, Vi_2 Vf_2 and are all equal. Phase 3: Puck slides on concrete. ———————————————– At this stage : Publication rate: Vi_3 =? Final speed: m Vf_3 = 24.9 / s Distance traveled: d3 =? Elapsed time: t3 =? Acceleration: a3 = -10.5 m / s ² Notes on Phase 3: * The initial rate of phase 3 (Vi_3) is the same as the speed of the end of phase 2 (Vf_2). * = We said Vf_2 Vi_2 Vf_1. This means that everyone is equal Vi_3 them. * The problem said Phase 3 ends "126 feet from the point of impact." This means that the total distance (D1 + D2 + D3) every 126 feet short as we know: (1) a1 = (Vf_1 – Vi_1) / t1 = Vf_1 / 0.0133 (2) d2 = 12.2 feet (3) = = = Vi_2 Vf_1 Vi_3 Vf_2 (4) d1 + d2 + d3 = 126 m (5) a3 = -10.5 m / s ² (6) Vf_3 = 24.9 m / s As we are given some information about the distance in phase 3, is an indication that we should use the kinematic equation for acceleration, speed and distance: 2 (a3) (D3) = (Vf_3) ² = (Vi_3) ² In our six previous equations can be written as: 2 (-10.5 m / s ²) (126ft – 12.2ft – d1) ² = (24.9 m / s) – (Vi_3) ² from the equation. 1 and Eq. 3, we get: Vi_3 = (a1) (0.0133). Replaces the previous 2 (-10.5 m / s ²) (113.8ft – d1) = (24.9 m / s) ² – ((a1) (0.0133)) ² This equation always has two unknown (A1 and D1). I do not see any way to get rid of the "D1". I think maybe because it refers to "the point of impact" which is supposed to assume that the distance traveled during contact negigibly disk is small, ie set d1 = 0. In this case we have: 2 (-10.5 m / s ²) (113.8ft) ² = (24.9 m / s) – ((a1) (0.0133)) ² Finally, just solved that "A1".