1. A plane accelerated from a speed of 2.0 m/s at the rate of 3.0 m/s2 over a distance of 530 m. What is its speed after traveling this distance?
2. A jet plane must achieve a velocity of 71 m/s for takeoff. From rest and with a 1,000 m long runway, what must the acceleration be?
3. The highest deceleration survived by a human (voluntary) is 452.76 m/s2. In 1954, Colonel John Stapp rode a rocket-powered sled that had a top speed of 632 mph. Colonel Stapp came to stop in how many meters?
4. A car decelerates at 8.0 m/s2 from 30 m/s to zero. How far does the car travel?
5. Scientists are developing new types of guns that might someday be used to launch satellites as if they were bullets. One such gun can give a small object a velocity of 3500 m/s moving through a distance of only 0.02 m. What acceleration does this gun give the object?
6. Highway safety engineers build soft barriers so that cars hitting them will slow down at a safe rate. A person wearing a shoulder harness can survive a car crash if the acceleration is smaller than -300 m/s2. Assuming constant acceleration, how thick should barriers be to stop car going 65 mph?
7. A baseball pitcher can throw a fastball at 44 m/s. The acceleration occurs as the pitcher holds the ball in his hand and moves the ball from rest through a distance of 3.5 m. Calculate the acceleration of the ball.
8. At an accident scene, 276-m long skids marks are found on the road. If the car that made these skid marks can decelerate at –9.8 m/s2, how fast was the car going before braking?
9. Time travel requires vehicles that can move at close to the speed of light, 3x108 m/s. Starting from rest, how far would such a vehicle have to travel if the acceleration is 20 m/s2?
Nobody will do your work for you, Ray.
But I will be glad to look over your answers.
1. To find the final speed of the plane after traveling a distance of 530 m, we can use the equation of motion:
Final velocity^2 = Initial velocity^2 + 2 * acceleration * distance
Given:
Initial velocity (u) = 2.0 m/s
Acceleration (a) = 3.0 m/s²
Distance (s) = 530 m
Substituting the values into the equation:
Final velocity^2 = (2.0 m/s)^2 + 2 * (3.0 m/s²) * (530 m)
Solving the equation, we get:
Final velocity^2 = 4.0 m²/s² + 3180 m²/s²
Final velocity^2 = 3184 m²/s²
Taking the square root of both sides, we find:
Final velocity = √3184 m/s ≈ 56.46 m/s
Therefore, the speed of the plane after traveling this distance is approximately 56.46 m/s.
2. To find the required acceleration for the jet plane to achieve a velocity of 71 m/s from rest over a distance of 1000 m, we can use the equation of motion:
Final velocity^2 = Initial velocity^2 + 2 * acceleration * distance
Given:
Initial velocity (u) = 0 m/s (as the plane starts from rest)
Final velocity (v) = 71 m/s
Distance (s) = 1000 m
Substituting the values into the equation:
(71 m/s)^2 = (0 m/s)^2 + 2 * acceleration * 1000 m
Simplifying the equation, we get:
5041 m²/s² = 2000 * acceleration
Dividing both sides by 2000, we find:
acceleration = 5041 m²/s² / 2000 = 2.5205 m/s²
Therefore, the acceleration required for the jet plane to achieve a velocity of 71 m/s over a distance of 1000 m is approximately 2.5205 m/s².
3. To calculate the distance required for Colonel John Stapp to come to a stop while decelerating at -452.76 m/s² from a speed of 632 mph, we need to convert the speed to meters per second and use the equation of motion:
Final velocity^2 = Initial velocity^2 + 2 * acceleration * distance
Given:
Initial velocity (u) = 632 mph
Acceleration (a) = -452.76 m/s²
Final velocity (v) = 0 m/s (as Colonel Stapp comes to stop)
We need to convert 632 mph to meters per second using the conversion factor: 1 mph = 0.44704 m/s.
Initial velocity (u) = 632 mph * 0.44704 m/s/mph = 282.816 m/s
Substituting the values into the equation:
(0 m/s)^2 = (282.816 m/s)^2 + 2 * (-452.76 m/s²) * distance
Simplifying the equation, we get:
0 m²/s² = 79914.084 m²/s² + (-905.52 m²/s²) * distance
Rearranging the equation, we find:
distance = -79914.084 m²/s² / (-905.52 m²/s²) = 88.255 m
Therefore, Colonel John Stapp would come to a stop in approximately 88.255 meters.
(Continued in the next message)