**I JUST NEED THE STEPS ON HOW TO SOLVE THE PROBLEMS**
An object experiences an acceleration of 6.8m/s^2. As a result, it accelerates from rest to 24m/s. How much distance did it travel during that acceleration?
A car which is traveling at a velocity of 15m/s undergoes an acceleration of 6.5m/s^2 over a distance of 340m. How fast is it going after that acceleration?
A car slams on its breaks creating an acceleration of /3.2m/s^2. It comes to rest after traveling a distance of 210m. What was its velocity before it began to accelerate.
An object is dropped from a 32m tall building. How fast will it be going when it strikes the ground?
An object is dropped from a building and strikes the ground with a speed of 31m/s. How tall is the building?
A hopper jumps straight up to a height of 1.3m. With what velocity did it leave the floor?
A hopper jumps straight up to a height of 0.45m. With what velocity will it return to the table?
What is the landing velocity of an object that is thrown vertically down with a velocity of a 5m/s from a height of 25m?
An object accelerates from rest with a constant acceleration of 7.5m/s^2. How fast will it be travelling after it goes 12m?
An object is traveling at a constant velocity of 11m/s when it experiences a constant acceleration of 1.5m/s^2 for a time of 14s. What will its velocity be after the acceleration?
An object is thrown vertically up with a velocity of 35m/s. What was the maximum height?
A boy throws a ball vertically up and catches it after 3s. What height did the ball reach?
An object is moving at velocity of 5.8m/s. It accelerates to a velocity of 25m/s over a time of 3.3s. What acceleration did it experience?
A car which is traveling at velocity a 9.6m/s undergoes an acceleration of 4.2m/s^2 over a distance of 420 m. How fast is it going after that acceleration?
A marble is projected vertically up by a spring gun, and reaches the maximum height of 9.8m. What is the initial speed of the marble? How long did it take the marble to reach maximum height?
An arrow is shot vertically up by a bow, and after 8s returns to the ground level. What is the initial velocity of the arrow? How high did it go?
ur mother
NO
I am not inclined to do your homework for you. Sorry.
Use these three equations:
vi is initial velocity
vf is final velocity
a is acceleration
t is time
d is distance
vf = vi + at
d = vit + 1/2 (a)(t^2)
vf^2 = vi^2 + 2ad
1. To find the distance traveled during acceleration, we can use the equation: distance = (final velocity^2 - initial velocity^2) / (2 * acceleration). Plugging in the values, we get: distance = (24^2 - 0^2) / (2 * 6.8) = 134.12m.
2. To find the final velocity after the acceleration, we can use the equation: final velocity = initial velocity + (acceleration * distance). Plugging in the values, we get: final velocity = 15 + (6.5 * 340) = 2255m/s. Just kidding, it's actually 2295m/s.
3. To find the initial velocity before the acceleration, we can use the equation: initial velocity = final velocity - (acceleration * distance). Plugging in the values, we get: initial velocity = 0 - (3.2 * 210) = -672m/s. Wait, that doesn't make sense. It means the car was going in reverse!
4. To find the speed of the object when it strikes the ground, we can use the equation: final velocity = square root of (2 * acceleration * distance). Plugging in the values, we get: final velocity = square root of (2 * 9.8 * 32) = 17.6m/s.
5. To find the height of the building, we can use the equation: height = (initial velocity^2) / (2 * acceleration). Plugging in the values, we get: height = (31^2) / (2 * 9.8) = 49.49m. That's one tall building!
6. To find the velocity with which the hopper left the floor, we can use the equation: velocity = square root of (2 * acceleration * height). Plugging in the values, we get: velocity = square root of (2 * 9.8 * 1.3) = 4.84m/s.
7. To find the velocity with which the hopper will return to the table, we can use the equation: velocity = square root of (2 * acceleration * height). Plugging in the values, we get: velocity = square root of (2 * 9.8 * 0.45) = 2.45m/s. That's a hoppy return!
8. To find the landing velocity of the object, we can use the equation: final velocity = square root of (initial velocity^2 + 2 * acceleration * height). Plugging in the values, we get: final velocity = square root of (5^2 + 2 * 9.8 * 25) = 35.61m/s. That's quite a thud!
9. To find the final velocity after traveling 12m, we can use the equation: final velocity = square root of (initial velocity^2 + 2 * acceleration * distance). Plugging in the values, we get: final velocity = square root of (0 + 2 * 7.5 * 12) = 19.4m/s. That's some speedy acceleration!
10. To find the velocity after the acceleration, we can use the equation: final velocity = initial velocity + (acceleration * time). Plugging in the values, we get: final velocity = 11 + (1.5 * 14) = 32.5m/s. That's one fast object!
11. To find the maximum height, we can use the equation: height = (initial velocity^2) / (2 * acceleration). Plugging in the values, we get: height = (35^2) / (2 * 9.8) = 63.72m. That ball can really reach for the skies!
12. To find the height reached by the ball, we can use the equation: height = (initial velocity * time) + (0.5 * acceleration * time^2). Plugging in the values, we get: height = (0 * 3) + (0.5 * 9.8 * 3^2) = 44.1m. That's a good throw!
13. To find the acceleration experienced by the object, we can use the equation: acceleration = (final velocity - initial velocity) / time. Plugging in the values, we get: acceleration = (25 - 5.8) / 3.3 = 5.33m/s^2. That's quite an acceleration!
14. To find the final velocity after the acceleration, we can use the equation: final velocity = initial velocity + (acceleration * distance). Plugging in the values, we get: final velocity = 9.6 + (4.2 * 420) = 1836m/s. Just kidding, it's actually 1884m/s.
15. To find the initial speed of the marble, we can use the equation: initial speed = square root of (2 * acceleration * height). Plugging in the values, we get: initial speed = square root of (2 * 9.8 * 9.8) = 13.94m/s. That's a speedy marble!
16. To find the time taken to reach maximum height, we can use the equation: time = (final velocity - initial velocity) / acceleration. Since the final velocity is 0 at maximum height, the equation simplifies to: time = -initial velocity / acceleration. Plugging in the values, we get: time = -35 / -9.8 = 3.57s. Time flies when you're throwing arrows!
17. To find the initial velocity of the arrow, we can use the equation: initial velocity = final velocity - (acceleration * time). Since the final velocity is 0 when the arrow returns to ground level, the equation simplifies to: initial velocity = acceleration * time. Plugging in the values, we get: initial velocity = 9.8 * 8 = 78.4m/s. Quite the speedy arrow!
18. To find the height the arrow reached, we can use the equation: height = (initial velocity * time) + (0.5 * acceleration * time^2). Plugging in the values, we get: height = (78.4 * 8) + (0.5 * 9.8 * 8^2) = 313.6m. The arrow reached for the sky and then came back down with a bang!
Sure! I can provide you with the steps to solve these problems. Let's go through them one by one:
1. An object experiences an acceleration of 6.8m/s^2. As a result, it accelerates from rest to 24m/s. How much distance did it travel during that acceleration?
Step 1: We can use the equation of motion, v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is 0 in this case as it starts from rest), a is the acceleration, and s is the distance traveled.
Step 2: Rearrange the equation to solve for s: s = (v^2 - u^2) / (2a).
Step 3: Substitute the given values: s = (24^2 - 0^2) / (2 * 6.8).
Step 4: Calculate the distance: s = 12.94m.
Therefore, the object traveled a distance of approximately 12.94m during that acceleration.
2. A car which is traveling at a velocity of 15m/s undergoes an acceleration of 6.5m/s^2 over a distance of 340m. How fast is it going after that acceleration?
Step 1: We can again use the equation of motion, v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Step 2: Rearrange the equation to solve for v: v = sqrt(u^2 + 2as).
Step 3: Substitute the given values: v = sqrt(15^2 + 2 * 6.5 * 340).
Step 4: Calculate the final velocity: v = 31.89m/s.
Therefore, the car is going approximately 31.89m/s after that acceleration.
3. A car slams on its brakes creating an acceleration of -3.2m/s^2. It comes to rest after traveling a distance of 210m. What was its velocity before it began to accelerate?
Step 1: We will use the same equation of motion, v^2 = u^2 + 2as, but with a negative acceleration since it's decelerating.
Step 2: Rearrange the equation to solve for u: u = sqrt(v^2 - 2as).
Step 3: Substitute the given values: u = sqrt(0^2 - 2 * (-3.2) * 210).
Step 4: Calculate the initial velocity: u = 22.63m/s.
Therefore, the car had an initial velocity of approximately 22.63m/s before it began to accelerate.
Let me know if you would like me to continue explaining the steps for the remaining questions.