If velocity is km/hr and time is in hours then acceleration will be in km/hr2
Suppose a projectile is fired from a gun, we know that "g" remains constant and as we use horizontal component of velocity in range sov0 also remains constant. Only sin2θ responsible for change in range. The range will be maximum if sin2θ has its maximum value that is 1.for maximum range:sin2θ = 12θ = sin-1 (1)θ = 90/2θ = 45 (degree)therefor if projectile is projected with the angle of 45(degree) its range will be maximum.
I suppose you should go back and find out where you made your mistake and redo it before any further mistakes are made! -apex
It suppose to keep cats off.
From's Newton's second law of motion:F=maor, F=m(v-u)/tor, F=(mv - mu)/tor,Ft=mv - muwhere F=force, m=mass, v=Initial Velocity, v=Final Velocity, t=timeThe above equation could be called Newton's first law of motion when u=v and F=0. So it could be concluded that a body of mass m would continue in a state of uniform motion(i.e it would move with Initial Velocity u) over a time t secondssuch that no external force acts on it(as F=0). When the body is at rest(i.e when u=v=0) then also it would remain at rest over a time t.By second law of motion, F is directly proportional to m, the mass and acceleration "a".F = k .m .awhere k is the constant of proportionality.a = F / k .m.when a = 0,F = 0 as m cannot be infinite.hence, if a body is at rest or in uniform motion, a = 0. thus the body remains in its state of rest or uniform motion.which is the first law.
3 weeks
Yes, an object's speed can still increase if its acceleration decreases, as long as the acceleration is positive. If the acceleration decreases but remains positive, the object will continue to speed up, just at a slower rate.
Let us suppose that the displacement is given by, x = kt2 , where k is constant of proportionality. Therefore, velocity of the body, v = dx/dt = d(kt2)/dt = 2kt Since, velocity depends on time ,the body is not moving with uniform velocity.... Again, acceleration of the body, a = dv/dt = d(2kt)/dt = 2k As the acceleration is independent of time , the body is moving with uniform acceleration..
Suppose v = [final] velocity, a = acceleration and s = distance thenif the initial velocity, u, is known,v^2 - u^2 = 2*a*sand if the elapsed time, t, is known,s = v*t - 0.5*a*t^2The subject of either equation may be changed to solve for the missing variable.
I'm trying to give you a simple example. Which hope will be able to make you understand that thing. We read physics in our language. So I can make some mistakes to write that in English... But I hope I won't be mistaken. By the way here it is... Suppose, There is a simple oscillator what is moving under the angles of 4 degrees. When it is in his height position then we know that it stops for a moment and then comes back to the equilibrium position. So in the height position it's velocity is 0 m/s. (cause it stops.) Suppose it's 1st velocity or the velocity in the equilibrium position is "v" and in that position the last velocity is 0. so we can calculate it's acceleration like that a = (last velocity-1st velocity)/time so a = (0-u)/t = - u/t, what is the maximum acceleration of a this body, (-) sign means the acceleration is gonna go down. But when it is coming back to the equilibrium position, it's velocity goes up.. And in the equilibrium position it's velocity is maximum. Then the velocity decreases again. so in the equilibrium position the velocity is maximum, in the equilibrium position which is "v". so it's acceleration will be a= v-v/t = 0/t = 0, that means the acceleration is zero when the velocity is maximum... That means the acceleration of a body can be zero when the velocity is maximum and the velocity can be zero when the acceleration is maximum. [Note: Always remember, to start to calculate from the equilibrium position. Because the oscillation starts from the equilibrium position. And what I said is the most simple statement. It can also be described by the equations of the a simple harmonic oscillations] - by JAS
Your initial velocity is 6 m/s, and your final velocity is 10 m/s. The total change in velocity is 4 m/s (10 m/s - 6 m/s). Given that this change occurs over 20 seconds, you can calculate the acceleration using the formula acceleration = change in velocity / time. Therefore, the acceleration is 0.2 m/s^2.
firstly, we should know what is meaning of accelerate i.e(to increase)that means acceleration is something which increases some physical quantity.In mechanics,acceleration means rate of change of velocity per unit time .a=v-u/twhere v=final velocity, u=initial velocity ,t=time& a=accelerationnow ,if the change in velocity is positive then i.e final velocity is greater than initial velocity.Suppose to be positive acceleration or simply acceleration.But when change in velocity is negative then acceleration is negative suppose to bedeceleration.Important thing is that acceleration is vector .so ,positive or negative accelerationindicates direction of acceleration.hence , force ,by Newton's second law.acceleration tends to increase the velocity of a moving body in the direction of motion while deceleration tends to decrease the velocity of the moving body applies in opposite direction of motion upto zero velocity after which it accelerate means increase velocity of the body but in opposite direction of initial direction of motion.
1. Gravity is acceleration. 2. speed is a scalar quantity. 3 . velocity is a vector quantity. In other words, gravity is the force which accelerates matter towards the centre of a mass. Speed is how fast something is moving in general. Velocity is speed in one direction. I suppose a link could be using the suvat equations with velocity and acceleration. But vector and scalar quantities don't mix very well.
acceleration doubles too.
acceleration is a relative quantity . state of rest or motion is also relative . if two body is in rest or moving with same velocity and having same acceleration then one is in state of rest with respect to other . suppose a person sitting in a train then he is in rest with respect to train but he is moving with the acceleration of train with respect to the ground.
velocity is a vector quantity and also acceleration is a vector quantity. Suppose my automobile is travelling in north direction and I apply brakes to it then until the automobile stops it will move in north direction while the acceleration will act in south direction as brakes are applied.
If the graph of distance traveled vs. time is not a straight line, it indicates that the object's acceleration is not constant. Acceleration is the rate of change of velocity, so a non-linear distance-time graph suggests that the object's velocity is changing at a non-constant rate, causing a curved graph.
Yes, it can, if the initial velocity vector of an object was in opposite direction to its constant acceleration. Example: Anything you toss with your hand has constant acceleration after you toss it ... the acceleration of gravity, directed downward. If you toss it upward, it starts out with upward velocity, which reverses and eventually becomes downward velocity.