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# 3 laws of uniformly accelerated motion

From Wikipedia

Acceleration

In physics, acceleration is the rate of change of velocity over time. In one dimension, acceleration is the rate at which something speeds up or slows down. However, since velocity is a vector, acceleration describes the rate of change of both the magnitude and the direction of velocity. Acceleration has the dimensionsL&nbsp;Tâˆ’2. In SI units, acceleration is measured in meters per second per second (m/s2).

Proper acceleration, the acceleration of a body relative to a free-fall condition, is measured by an instrument called an accelerometer.

In common speech, the term acceleration is used for an increase in speed (the magnitude of velocity); a decrease in speed is called deceleration. In physics, a change in the direction of velocity also is an acceleration: for rotary motion, the change in direction of velocity results in centripetal (toward the center) acceleration; where as the rate of change of speed is a tangential acceleration.

In classical mechanics, for a body with constant mass, the acceleration of the body is proportional to the net force acting on it (Newton's second law):

where F is the resultant force acting on the body, m is the mass of the body, and a is its acceleration.

## Average and instantaneous acceleration

Average acceleration is the change in velocity (Î”'v) divided by the change in time (Î”t). Instantaneous acceleration is the acceleration at a specific point in time which is for a very short interval of time as Î”t approaches zero.

The velocity of a particle moving on a curved path as a function of time can be written as:

\mathbf{v} (t) =v(t) \frac {\mathbf{v}(t)}{v(t)} = v(t) \mathbf{u}_\mathrm{t}(t) ,

with v(t) equal to the speed of travel along the path, and

\mathbf{u}_\mathrm{t} = \frac {\mathbf{v}(t)}{v(t)} \ ,

a unit vector tangent to the path pointing in the direction of motion at the chosen moment in time. Taking into account both the changing speed v(t) and the changing direction of ut, the acceleration of a particle moving on a curved path on a planar surface can be written using thechain rule of differentiation and the derivative of the product of two functions of time as:

\begin{alignat}{3}

\mathbf{a} & = \frac{d \mathbf{v}}{dt} \\ & = \frac{\mathrm{d}v }{\mathrm{d}t} \mathbf{u}_\mathrm{t} +v(t)\frac{d \mathbf{u}_\mathrm{t}}{dt} \\ & = \frac{\mathrm{d}v }{\mathrm{d}t} \mathbf{u}_\mathrm{t}+ \frac{v^2}{R}\mathbf{u}_\mathrm{n}\ , \\ \end{alignat}

where un is the unit (inward) normal vector to the particle's trajectory, and R is its instantaneous radius of curvature based upon the osculating circle at time t. These components are called the tangential accelerationand the radial acceleration or centripetal acceleration (see alsocircular motion and centripetal force).

Extension of this approach to three-dimensional space curves that cannot be contained on a planar surface leads to the Frenet-Serret formulas.

## Special cases

### Uniform acceleration

Uniform or constant acceleration is a type of motion in which the velocity of an object changes by an equal amount in every equal time period.

A frequently cited example of uniform acceleration is that of an object in free fall in a uniform gravitational field. The acceleration of a falling body in the absence of resistances to motion is dependent only on the gravitational field strength g (also called acceleration due to gravity). By Newton's Second Law the force, F, acting on a body is given by:

\mathbf {F} = m \mathbf {g}

Due to the simple algebraic properties of constant acceleration in the one-dimensional case (that is, the case of acceleration aligned with the initial velocity), there are simple formulae that relate the following quantities: displacement, initial velocity, final velocity, acceleration, and time:

\mathbf {v}= \mathbf {u} + \mathbf {a} t
\mathbf {s}= \mathbf {u} t+ \over {2}} \mathbf {a}t^2 = \over {2}}

where

\mathbf{s} = displacement
\mathbf{u} = initial velocity
\mathbf{v} = final velocity
\mathbf{a} = uniform acceleration
t = time.

In the case of uniform acceleration of an object that is initially moving in a direction not aligned with the acceleration, the motion can be resolved into two orthogonal parts, one of constant velocity and the other according to the above equations. As Galileo showed, the net result is parabolic motion, as in the trajectory of a cannonball, neglecting air resistance.

### Circular motion

An example of a body experiencing acceleration of a uniform magnitude but changing direction is uniform <

Question:help...........

Answers:v = u + at s = 1/2 * (u+v)t s = ut + 1/2 at^2 v^2 = u^2 + 2as can find in any basic textbook...

Question:If a ball falls 10 meters in 2 seconds, what is the acceleration? and if you drop a ball on an inclined plane 3 times and square the average and make a graph of it...what does the shape of the distance vs time^2 graph indicate?

Answers:the basic equation is: dist = 1/2 a * t^2 a = 2 * dist / t^2 = 20 / 4 = 5 m/sec^2

Question:A box accidentally drops from a truck traveleing 13.5 m/s and slides along the ground for a distance of 30.5 meters decelerating uniformly the entire way. Find: 1. The negative acceleration 2.The time for the box to come to rest 3. The distance traveleed during the first second. So i answered these but im not sure if im correct so thanks if you can help !

Answers:v = (v0) + 2ax 0 = (13.5) + 2a(30.5) a = -13.5 / 2(30.5) a = -2.99 m/s <=============== x = x0 + (v0)t + at 30.5 = (0) + 13.5t + (-2.99)t -1.49t + 13.5t - 30.5 = 0 The Quadratic Formula: If ax + bx + c = 0 Then x = { -b (b - 4ac) } / 2a t = 4.51 s <=================== x = (v0)t + at x = 13.5(1) + (-2.99) (1) x = 12.0 m <===================

Question:Determine the stopping distances of an automobile with an initial speed of 90 km/h and human reaction time of 1.0s.

Answers:First convert the speed into SI units. 90 Km/hr =90000m/3600 sec =25 m/s So the car will travel 25m BEFORE the human gets near the brake. The rest of the stopping distance depends on the maximum friction obtainable at the rubber of the wheels. This can be as high as 10m/s^2 for good tyres on dry bitumen or as low as 3m/s^2 on a wet slippery road But it can be well under 1m/s^2 on icy surfaces. Without specifying the friction available then we can only give various possibilities. For the first scenario we get v^2 = 2* a *s or s= v^2/(2*a) =25*25/20 =32 m approximately So in this case the total stopping distance =57 m If a typical car is no more than 5m long this is 11 or twelve car lengths.!! No wonder so many people have rear end collisions on freeways. Doing the same for ice we get a stopping distance of 320 m (on a level road) Adding the reaction distance gives a total stopping distance of 345 m.