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# definition of linear motion acceleration

From Wikipedia

Linear motion

Linear motion is motion along a straight line, and can therefore be described mathematically using only one spatial dimension. It can be uniform, that is, with constant velocity (zero acceleration), or non-uniform, that is, with a variable velocity (non-zero acceleration). The motion of a particle (a point-like object) along the line can be described by its position x, which varies with t (time). Linear motion is sometimes called rectilinear motion.

An example of linear motion is that of a ball thrown straight up and falling back straight down.

The average velocity v during a finite time span of a particle undergoing linear motion is equal to

v = \frac {\Delta d}{\Delta t}.

The instantaneous velocity of a particle in linear motion may be found by differentiating the position x with respect to the time variable t. The acceleration may be found by differentiating the velocity. By the fundamental theorem of calculus the converse is also true: to find the velocity when given the acceleration, simply integrate the acceleration with respect to time; to find displacement, simply integrate the velocity with respect to time.

This can be demonstrated graphically. The gradient of a line on the displacement time graph represents the velocity. The gradient of the velocity time graph gives the acceleration while the area under the velocity time graph gives the displacement. The area under an acceleration time graph gives the velocity.

Linear motion is the most basic of all motions. According to Newton's first law of motion, objects not subjected to forces will continue to move uniformly in a straight line indefinitely. Under every-day circumstances, external forces such as gravity and friction will cause objects to deviate from linear motion and can cause them to come to a rest.

For linear motion embedded in a higher-dimensional space, the velocity and acceleration should be described as vectors, made up of two parts: magnitude and direction. The direction part of these vectors is the same and is constant for linear motion, and only for linear motion

Motion graphs and derivatives

In mechanics, the derivative of the position vs. timegraph of an object is equal to the velocity of the object. In the International System of Units, the position of the moving object is measured in meters relative to the origin, while the time is measured in seconds. Placing position on the y-axis and time on the x-axis, the slope of the curve is given by:

v = \frac{\Delta y}{\Delta x} = \frac{\Delta s}{\Delta t}.

Here s is the position of the object, and t is the time. Therefore, the slope of the curve gives the change in position (in metres) divided by the change in time (in seconds), which is the definition of the average velocity (in meters per second (\begin{matrix} \frac{m}{s} \end{matrix})) for that interval of time on the graph. If this interval is made to be infinitesimally small, such that {\Delta s} becomes {ds} and {\Delta t} becomes {dt}, the result is the instantaneous velocity at time t, or the derivative of the position with respect to time.

A similar fact also holds true for the velocity vs. time graph. The slope of a velocity vs. time graph is acceleration, this time, placing velocity on the y-axis and time on the x-axis. Again the slope of a line is change in y over change in x:

a = \frac{\Delta y}{\Delta x} = \frac{\Delta v}{\Delta t}.

Where v is the velocity, measured in \begin{matrix} \frac{m}{s} \end{matrix}, and t is the time measured in seconds. This slope therefore defines the average acceleration over the interval, and reducing the interval infinitesimally gives \begin{matrix} \frac{dv}{dt} \end{matrix}, the instantaneous acceleration at time t, or the derivative of the velocity with respect to time (or the second derivative of the position with respect to time). The units of this slope or derivative are in meters per second per second (\begin{matrix} \frac{m}{s^2} \end{matrix}, usually termed "meters per second-squared"), and so, therefore, is the acceleration.

Since the acceleration of the object is the second derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object. (Velocity is on the y-axis and time on the x-axis. Multiplying the velocity by the time, the seconds cancel out and only meters remain. \begin{matrix} \frac{m}{s} \end{matrix}s = m.)

The same multiplication rule holds true for acceleration vs. time graphs. When (\begin{matrix} \frac{m}{s^2} \end{matrix}) is multiplied by time (s), velocity is obtained. (\begin{matrix} \frac{m}{s^2} \end{matrix}s = \begin{matrix} \frac{m}{s} \end{matrix}).

## Variable rates of change

The expressions given above apply only when the rate of change is constant or when only the average (mean) rate of change is required. If the velocity or positions change non-linearly over time, such as in the example shown in the figure, then differentiation provides the correct solution. Differentiation reduces the time-spans used above to be extremely small and gives a velocity or acceleration at each point on the graph rather than between a start and end point. The derivative forms of the above equations are

v = \frac{ds}{dt},
a = \frac{dv}{dt}.

Since acceleration differentiates the expression involving position, it can be rewritten as a second derivative with respect to position:

a = \frac{d^2 s}{dt^2}.

Since, for the purposes of mechanics such as this, integration is the opposite of differentiation, it is also possible to express position as a function of velocity and velocity as a function of acceleration. The process of determining the area under the curve, as described above, can give the displacement and change in velocity over particular time intervals by using definite integrals:

s(t_2)-s(t_1) = \int_{t_1}^{t_2}{v}\, dt,
v(t_2)-v(t_1) = \int_{t_1}^{t_2}{a}\, dt.

Question:Please give me the definition of the uniformly accelerated linear motion. Thanks.

Answers:Motion implies momentum, which implies velocity. Linear implies a straight line. Accelerating implies changing velocity. And uniform implies constancy. So, when a body moves in a straight line and accelerates at a constant rate, the body is said to have an uniformly accelerated linear motion.

Question:1. batman is sitting in the batmobile at a spotlight. As the light turns green, Robin passes Batman in his lime green pinto at a constant speed of 60km/h. If batman gives chase, accelerating at constant rate of 10km/h/s, determine am completely off on my answers a. how long it takes batman to attain same speed? answer :6.0 b. how far Batman travels in this time? ans: 50m c. how long it takes for Batman to catch up to robin? ans : 12.0s 2. A squash ball makes contact with a squash racquet and changes from 15m/s to 25m/s east in 0.10s. determine the vector acceleration of the squash ball

Answers:a. The units as given are OK for this one. The speed changes by 10 km/h each second, and he wants to get to 60 km/h. So 60/10 = 6 seconds. b. You can use d = (1/2)at^2 for this, but to get an answer in meters you're going to have to convert the acceleration to m/sec^2. To do that, convert the km to meters (multiply by 1000) and the hr to seconds (divide by 3600). c. You can use d = (1/2)at^2 again, and again you need the value of a in m/sec^2. This time t is unknown. It is going to be more than 6 seconds and Batman is going to continue to accelerate (he won't catch Robin if he stops accelerating when he gets to 60, because Robin is far ahead of him, also doing 60). In the same time, Robin goes d = v*t, but you're going to have to convert his speed to m/sec also. Batman catches Robin when these distances are the same. (1/2)at^2 = v*t You can divide out one factor of t (1/2)at = v and you can rearrange that to solve for v. Actually come to think of it, you can use the original units of km/hr for v and km/hr/sec for a, since they are compatible in this equation. You've got km/hr on both sides so that would be OK. Number 2, just use the definition of acceleration. You have the change in velocity. You have the time. a = (change in velocity)/time

Question:Can you help me with this question, I keep getting time=2sec which isn't right. P and Q are points 162m apart. A body leaves P with initial speed 5m/s and travels towards Q with uniform acceleration 3m/s^2. At the same instant another body leaves Q and travels towards P with initial speed 7m/s and uniform acceleration 2m/s^2. After how many seconds do they meet and what, then is the speed of each body? The answers are: 6 s, 23m/s, 19m/s. But I can't work it out.

Answers:1 st use distance covered s=ut+1/2 at^2 find s1 where u= initial velocity =5 and a=acceleration=3 thus s1=5t+(1/2)x3t^2 similarly find s2=7t+(1/2)x2t^2 now since they meet, s1+s2=162 substituting s1 and s2 in above eq we get a quadratic eq as below 5t^2+24t-324=0 solving it we get 2 values t=6 or -10.8 negative value of time is not possible thus we take t=6 seconds now use final velocity v=u+at thus v1=5+3x6=23 and v2=7+2x6=19 hope this helps

Question:The driver of a car travelling at 20m/s sees a second car 120m in front, travelling in the same direction at a uniform speed of 8m/s. (a) What is the least uniform retardation that must be applied to the faster car so as to avoid a collision? (b) If the actual retardation is 1m/s^2, calculate: (i) the time interval in seconds for the faster car to reach a point 66m behind the slower. (ii) the shortest distance between the cars. Please explain this to me fully. Thank you.

Answers:qa the RELATIVE velocity between the two cars = 20 - 8 = 12 m/s this must be brought down to zero while bridging the 120 m gap thus 12^2 = 2a*120 or a = 144/240 = 0.6 m/s^2 qb(i) actual retardation = 1 m/s^2 if faster car reaches 66m behind the slower, gap has been bridged by 120 - 66 = 54 m let the RELATIVE speed of the faster car reduce to v m/s, then 12^2 - v^2 = 2a*54 = 2*1*54 = 108 or v^2 = 144 - 108 = 36 so v = 6 m/s now distance = average velocity*time, so 54 = 0.5(12 + 6)t or t = 54/9 = 6 s qb(ii) minimum distance between the two will be when the faster car reduces relative speed to zero so 12^2 = 2d where d is the REDUCTION in distance between the two cars so d = 72 ft reduction from120 m and minimum distance between the two cars = 120 - 72 = 48 m ans: qa: 0.6 m/s^2 qb(i) 6s qb(ii) 48 m edit: note how simple the calculations hv become by considering RELATIVE velocity rather than ACTUAL velocities..