A motorbike increases its velocity from 20.0m/s(W) to 30.0m/s(W) over a distance of 200 m. What is the acceleration and time?
Start by using v² = v₀² + 2ax. Solving for a gives a = (v² - v₀²)/(2x). v = 30.0 m/s; v₀ = 20.0 m/s; x = 200 m, so a = ((30.0 m/s)² - (20 m/s)²) / (2(200 m)) = 1.25 m/s². Now use v = v₀ + at. t = (v - v₀)/a = ((30.0 m/s) - (20.0 m/s)) / (1.25 m/s²) = 8.00 s.
A cylinder is rolling in a plane with initial velocity 10 m/s, what distance will it travel?
The acceleration (in magnitude) as experienced by the cylinder is,a=N/m=ugWhere u- coefficient of kinetic friction,N- normal reaction offered by the plane.As the cylinder starts its journey with a velocity of 10 m/s and comes to rest after covering x (say) metres, we can useW^2 = V^2 + 2axWhere W- final velocity (w=0m/s)V- initial velocity.Now just plug in the values and juggle them a little bit.0 = (10 m/s)^2 + 2(-0.5*10 m/s^2)xx = 100 (m/s)^2 / 10 (m/s^2) = 10 m.The cylinder, as it turns out, goes no farther than 10 metres. And this took it 2 seconds to do that.
A cylinder is rolling in a plane with initial velocity 10 m/s, what distance will it travel?
The acceleration (in magnitude) as experienced by the cylinder is,a=N/m=ugWhere u- coefficient of kinetic friction,N- normal reaction offered by the plane.As the cylinder starts its journey with a velocity of 10 m/s and comes to rest after covering x (say) metres, we can useW^2 = V^2 + 2axWhere W- final velocity (w=0m/s)V- initial velocity.Now just plug in the values and juggle them a little bit.0 = (10 m/s)^2 + 2(-0.5*10 m/s^2)xx = 100 (m/s)^2 / 10 (m/s^2) = 10 m.The cylinder, as it turns out, goes no farther than 10 metres. And this took it 2 seconds to do that.
A speedboat increases its speed uniformly from 20 m/s to 30 m/s in a distance of 200m....find...?
Here's what we know: initial velocity (u) = 20 final velocity (v) = 30 distance (s) = 200 Now we can apply an equation of straight line motion: v^2 = u^2 + 2as The only thing we don't know is acceleration 'a'. 30^2 = 20^2 + 2(a)(200) 900 = 400 + 400a 500 = 400a a = 1.25 ms^(-2) To find the time we'll use another equation for straight line motion: v = u +at 30 = 20 + 1.25t 10 = 1.25t t = 8 seconds
Help with physics problems?? Speed, Distance, and Acceleration?
6. A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system, and comes to rest 5.5 s later. Find how far the car moves while stopping. 7. A car accelerates uniformly from rest to a speed of 23.7 km/h in 6.5 s. Find the distance in m the car travels during this time. 8. When Maggie applies the brakes of her car, the car slows uniformly from 15.00 m/s to 0.00 m/s in 2.50 s. How many meters before a stop sign must she apply her brakes in order to stop at the sign? 9. A jet plane lands with a velocity of 100. m/s and can accelerate at a maximum rate of -5.0 m/s2 as it comes to rest. Can this plane land at an airport where the runway is 0.80 km long? 10. A driver in a car traveling at a speed of 78 km/h sees a cat 100. m away on the road. How long will it take for the car to accelerate constantly to a stop in exactly 99 m? 11. A car enters the freeway with a speed of 23 km/h and accelerates to a speed of 86 km/h in 3.5 min. How far does the car move while accelerating? 12. A plane starting at rest at one end of a runway undergoes a constant acceleration of 4.8 m/s2 for 15 s before takeoff. What is its speed at takeoff? How long must the runway be for the plane to be able to take off? 13. A car with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s2 for 3.6 s. Find the final speed and the distance traveled of the car during this time. 14. An automobile with an initial speed of 4.30 m/s accelerates at the rate of 3.00 m/s2. Find the final speed and the distance travel after 5.0 s. 15. A car starts from rest and travels for 5.0 s with a uniform acceleration of 1.5 m/s2. What is the final velocity of the car? How far does the car travel in this time interval?
What is the difference between instantaneous and average velocity?
For this I'm going to use driving as an example. Imagine you're in a car do m going down the interstate going on a 60 mile trip. Depending on traffic, you could be going anywhere between 0 and 80 mi/h. Say that there is light traffic and you look at the speedometer and it says that you are traveling at 45 mi/h, that is your instantaneous velocity. Five minutes later, you look over an you are traveling at 80 mi/h, that is another instantaneous velocity. Instantaneous velocity is your velocity at any one time. Now let's say the entire trip took one hour so that is 60 mi divided by one hour, so your average velocity was 60 mi/h, even though at some points you were traveling faster or slower than 60 mi/h.
A particle moves along the x-axis, its position at time t given by x(t) = t/(1+t^2), t > or equal to 0?
James' answer is incorrect. jmurphy/odu83 is on the right track, but got the wrong expression for the second derivative (the acceleration), and he uses the wrong logic for determining when the particle is speeding up or slowing down (and his answer is therefore incorrect, and would be so even if his expression for the acceleration were correct). a) The particle is moving to the right whenever the velocity is positive, and moving to the left whenever the velocity is negative. The velocity of the particle is given by the derivative of x(t) with respect to time: v(t) = dx(t)/dt = (1-t^2)/((1+t^2)^2) which is the same expression as odu83 obtained. The only positive value of t for which this expression is equal to zero is t = 1. For 0<=t < 1, the velocity is positive, and the particle is moving to the right. At t = 1, the particle is momentarily at rest (reversing direction), and for t > 1, the velocity is negative, and the particle is moving to the left. b) For part (b), you can do as odu83 has done, and integrate the absolute value of the velocity, but there is a simpler way. We know that from t = 0 to t = 1, the particle is always moving to the right, reaching a maximum displacement of x = 0.5 at t = 1 (found by simply plugging t = 1 into the expression for x(t)). Then, from t = 1 to t = 4, the particle moves from x = 0.5 to x = 0.235, or a distance of (0.5 - 0.235) = 0.265, so the total distance is the sum of 0.5 + 0.265 = 0.765 (c), the acceleration is given by the second derivative of x(t) with respect to time. The correct expression for this is: a(t) = dv(t)/dt = 2*t*(t^2 - 3)/(1 + t^2)^3 The acceleration is zero at t = 0 and again at t = sqrt(3). The acceleration is negative for 0 < t < sqrt(3), and positive for t > sqrt(3). (d) The particle will be slowing down whenever the acceleration and velocity have opposite signs, and speeding up whenever the acceleration and velocity have the same sign (i.e, slowing when the acceleration and velocity vectors are opposed, and increasing speed when they are aligned). Summarizing the signs of v and a for various ranges of t: 0 < t < 1, v>0, a<0, (Particle slowing) 1 < t < sqrt(3), v< 0, a<0, (Particle speeding up) sqrt(3) < t, v<0, a>0, (Particle slowing down)
What is the difference between instantaneous and average velocity?
For this I'm going to use driving as an example. Imagine you're in a car do m going down the interstate going on a 60 mile trip. Depending on traffic, you could be going anywhere between 0 and 80 mi/h. Say that there is light traffic and you look at the speedometer and it says that you are traveling at 45 mi/h, that is your instantaneous velocity. Five minutes later, you look over an you are traveling at 80 mi/h, that is another instantaneous velocity. Instantaneous velocity is your velocity at any one time. Now let's say the entire trip took one hour so that is 60 mi divided by one hour, so your average velocity was 60 mi/h, even though at some points you were traveling faster or slower than 60 mi/h.
As a particle moves along the x axis it is acted upon by a single conservative force given by Fx = (20 - 4.0x)
ANSWER IS c. -18 J Fx = (20 - 4.0x) N but F= - dU/dx So - dU/dx =(20 - 4.0x) dU = 4.0 x dx - 20 dx Now integrate U = 2x^2 -20x + C at the origin x =0 U = +30 J 30 =C So the PE function is U= 2x^2 -20x +30 at x =+4 m U = 2(16)-80+30 U= - 18 Joules remember that If a force acting on an object is a function of position only, it is said to be a conservative force, and it can be represented by a potential energy function which for a one-dimensional case satisfies the derivative condition F= - dU/dx I hope that is worth 5 stars and you ace the test!
Questions please....?
Challenge: Two Objects are set into motion simultaneously. Object A is dropped from the top of a building. Object B is launched upward from the ground with an initial speed that is equal to Object A’s final speed. How long does it take for the objects to pass each other? At which position relative to the building do they pass each other?