(One-Dimensional Motion) A race car can accelerate from 0 to 100 mi/h (1mi = 1600 m) in 14 seconds, and its brake can provide a maximum deceleration of 0.65g (g = 9.8 m/s?). The driver wants to test the performance of the car by finding the shortest possible time it would take for the car to travel a distance of 0.5 mi in a straight line, starting from rest and finishing at rest. (a) What is the acceleration of the car in terms of g? (b) For how long should the driver be accelerating before reaching the top speed of the test drive?
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- A soccer ball starts from rest and accelerates with an acceleration of 0.385 m/s2 while moving down a 8.50 m long inclined plane. When it reaches the bottom, the ball rolls up another plane, where, after moving 14.75 m, it comes to rest. (a) What is the speed of the ball at the bottom of the first plane (in m/s)? (Round your answer to at least two decimal places.) m/s (b) How long does it take to roll down the first plane (in s)? (c) What is the magnitude of the acceleration along the second plane (in m/s?)? m/s2 (d) What is the ball's speed 8.20 m along the second plane (in m/s)? m/s (e) What If? You change the angles of the two inclines in such a manner that the new acceleration down the first incline is a, new down = 2a initiel down and up the second incline is a, new un x initial up, What is the ball's speed 8.20 m up the second plane (in m/s)? 2 m/shere is my problem, I need help: One simple model for a person running the 100 mm dash is to assume the sprinter runs with constant acceleration until reaching top speed, then maintains that speed through the finish line. If a sprinter reaches his top speed of 11.1 m/sm/s in 2.64 ss , what will be his total time? Express your answer in seconds.A hippopotamus can run up to 9.86 m/s. Suppose a hippopotamus uniformly accelerates 0.28 m/s² until it reaches a top speed of 9.86 m/s. If the hippopotamus has run 25.1 m, what is its initial speed in m/s? (round your answer to two decimal places; DO NOT include units)
- A race car can accelerate from 0 to 100 mi/h (1mi = 1600 m) in 14 seconds, and its brake can provide a maximum deceleration of 0.65g (g = 9.8 m/s²). The driver wants to test the performance of the car by finding the shortest possible time it would take for the car to travel a distance of 0.5 mi in a straight line, starting from rest and finishing at rest. (a) What is the acceleration of the car in terms of g? (b) For how long should the driver be accelerating before reaching the top speed of the test drive?A basketball starts from rest and accelerates with an acceleration of 0.395 m/s2 while moving down a 9.25 m long inclined plane. When it reaches the bottom, the ball rolls up another plane, where, after moving 15.00 m, it comes to rest. (a)What is the speed of the ball at the bottom of the first plane (in m/s)? (Round your answer to at least two decimal places.) m/s (b)How long does it take to roll down the first plane (in s)? s (c)What is the magnitude of the acceleration along the second plane (in m/s2)? m/s2 (d)What is the ball's speed 8.10 m along the second plane (in m/s)? m/sA startled armadillo leaps upward, rising 0.586 m in the first 0.204 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.586 m? (c) How much higher does it go? Use g=9.81 m/s2.
- From rest, the cheetah can accelerate at 8.8 m/s2 and reach a top speed of 30 m/s (108 km/h)! It can maintain this maximum speed over a distance of about 400 meters before it needs to stop.On the other hand, the Thomson’s gazelle has a top speed of 70 km/h, which is less than the cheetah’s, but it can maintain this top speed for a while as well. From rest, the gazelle can accelerate at 4.5 m/s2 to reach its top speed. Q1. After what total distance from rest must the cheetah stop? (clearly show calculations) Q2. In the elapsed time that the cheetah started and must stop, what distance can the gazelle cover? (again clearly show your calculations)From rest, the cheetah can accelerate at 8.8 m/s2 and reach a top speed of 30 m/s (108 km/h)! It can maintain this maximum speed over a distance of about 400 meters before it needs to stop.On the other hand, the Thomson’s gazelle has a top speed of 70 km/h, which is less than the cheetah’s, but it can maintain this top speed for a while as well. From rest, the gazelle can accelerate at 4.5 m/s2 to reach its top speed. When a cheetah goes after a gazelle, success or failure is a simple matter of kinematics: you will determine if the cheetah’s high speed is enough to allow it to reach its prey before it runs out of steam (or time).You will apply basic kinematics and simple assumptions to determine how a chase can play out. The Scene:A cheetah has spotted a gazelle. At the same instant the cheetah leaps into action, the gazelle has spotted him and starts heading directly away.Q1. After what total distance from rest must the cheetah stop? (clearly show calculations and substitutions) Q2.…A tortoise can run with a speed of 0.15 m/s, and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for 1.0 minutes. The tortoise wins by a shell (25 cm). How long does the race take? (sec) What is the length of the race?(m)
- From rest, the cheetah can accelerate at 8.8 m/s2 and reach a top speed of 30 m/s (108 km/h)! It can maintain this maximum speed over a distance of about 400 meters before it needs to stop.On the other hand, the Thomson’s gazelle has a top speed of 70 km/h, which is less than the cheetah’s, but it can maintain this top speed for a while as well. From rest, the gazelle can accelerate at 4.5 m/s2 to reach its top speed. When a cheetah goes after a gazelle, success or failure is a simple matter of kinematics: you will determine if the cheetah’s high speed is enough to allow it to reach its prey before it runs out of steam (or time).You will apply basic kinematics and simple assumptions to determine how a chase can play out. Q2. In the elapsed time that the cheetah started and must stop, what distance can the gazelle cover? (again clearly show your calculations) Why is the 70 multiplied by 5/18?Frogs, with their long, strong legs, are excellent jumpers. And thanks to the good folks of Calaveras County, California, who have a jumping frog contest every year in honor of a Mark Twain story, we have very good data as to just how far a determined frog can jump. The current record holder is Rosie the Ribeter, a bullfrog that made a leap of 2.2 m from a standing start. This compares favorably with the world record for a human, which is a mere 3.7 m. Typical data for a serious leap by a bullfrog look like this: The frog goes into a crouch, then rapidly extends its legs by 15 cm as it pushes off, leaving the ground at an angle of 30° to the horizontal. It’s in the air for 0.49 s before landing at the same height from which it took off. Given this leap, what is the acceleration while the frog is pushing off? How far does the frog jump?A tortoise can run with a speed of 0.12 m/s, and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for 4.0 minutes. The tortoise wins by a shell (35 cm). (a) How long does the race take?(b) What is the length of the race?