Kinematic Equations For Uniformly Accelerated Motion
For uniformly accelerated motion, we can derive some simple equations that relate displacement (x), time taken (t), initial velocity ($v_0$), final velocity (v) and acceleration (a). Equation (3.6) already obtained gives a relation between final and initial velocities v and $v_0$ of an object moving with uniform acceleration a:
$v=v_0+at$ (1)
This relation is graphically represented in Fig.1. The area under this curve is:
Area between instants 0 and t = Area of triangle ABC + Area of rectangle OACD
= $\frac{1}{2}(v-v_0)t+v_0t$
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| Fig. 1: Area under v-t curve for an object with uniform acceleration. |
As explained in the previous section, the area under v-t curve represents the displacement. Therefore, the displacement x of the object is:
$x=\frac{1}{2}(v-v_0)t+v_0t$ (2)
But $v-v_0=at$
Therefore, $x=\frac{1}{2}at^2+v_0t$
or, $x=v_0t+\frac{1}{2}at^2$ (3)
Equation (2) can also be written as
$x=\frac{(v_0+v)t}{2}=\bar{v}t$ (4a)
Where,
$\bar{v}=\frac{(v_0+v)t}{2}$ (constant acceleration only) (4b)
Equations (4a) and (4b) mean that the object has undergone displacement x with an average velocity equal to the arithmetic average of the initial and final velocities.
From Eq. (1), t = (v – $v_0$)/a. Substituting this in Eq. (4a), we get
$x=\bar{v}t=\left(\frac{v_0+v}{2}\right)\left(\frac{v-v_0}{a}\right)=\frac{v^2-v_0^2}{2a}$
$v^2=v_0^2+2ax$ (5)
This equation can also be obtained by substituting the value of t from Eq. (1) into Eq. (3). Thus, we have obtained three important equations:
$v=v_0+at$
$x=v_0t+\frac{1}{2}at^2$
$v^2=v_0^2+2ax$ (6a)
connecting five quantities $v_0$, v, a, t and x. These are kinematic equations of rectilinear motion for constant acceleration.
The set of Eq. (6.a) were obtained by assuming that at t = 0, the position of the particle, x is 0. We can obtain a more general equation if we take the position coordinate at t = 0 as non-zero, say $x_0$. Then Eqs. (6a) are modified (replacing x by x – $x_0$) to:
$v=v_0+at$
$x=x_0+v_0t+\frac{1}{2}at^2$
$v^2=v_0^2+2a(x-x_0)$ (6b)
Example 1
Obtain equations of motion for constant acceleration using method of calculus.
Answer
By definition $a =\frac{dv}{dt}$
$dv=adt$
Integrating both sides
$\int_{v_0}^{v}dv=\int_{0}^{t}adt$
$\int_{v_0}^{v}dv=a\int_{0}^{t}dt$
$v-v_0=at$
$v=v_0+at$
Further, $v=\frac{dx}{dt}$
$dx=vdt$
Integrating both sides
$\int_{x_0}^{x}dx=\int_{0}^{t}vdt$
$\int_{x_0}^{x}dx=\int_{0}^{t}(v_0+at)dt$
$x-x_0=v_0t+\frac{1}{2}at^2$
$x=x_0+v_0t+\frac{1}{2}at^2$
We can write
$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$
or $vdv=adx$
Integrating both sides,
$\int_{v_0}^{v}vdv=\int_{x_0}^{x}adx$
$\int_{v_0}^{v}dv=a\int_{0}^{t}dt$
$\frac{v^2-v_0^2}{2}=a(x-x_0)$
$v^2=v_0^2+2a(x-x_0)$
The advantage of this method is that it can be used for motion with non-uniform acceleration also.
Now, we shall use these equations to some important cases.
Example 1
A ball is thrown vertically upwards with a velocity of 20 m/s from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how long will it be before the ball hits the ground? Take g = 10 m.$s^{–2}$.
Answer
(a) Let us take the y-axis in the vertically upward direction with zero at the ground, as shown in Fig. 2.
Now $v_o$ = + 20 m$s^{–1}$,
a = – g = –10 m$s^{–2}$,
v = 0 m$s^{–1}$
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| Fig.2 |
If the ball rises to height y from the point of launch, then using the equation
$v^2=v_0^2+2a(y-y_0)$
we get
0 = (20)2 + 2(–10)(y – $y_0$)
Solving, we get, (y – $y_0$) = 20 m.
(b) We can solve this part of the problem in two ways. Note carefully the methods used.
FIRST METHOD: In the first method, we split the path in two parts: the upward motion (A to B) and the downward motion (B to C) and calculate the corresponding time taken $t_1$ and $t_2$. Since the velocity at B is zero, we have:
v = $v_o$ + at
0 = 20 – 10$t_1$
Or, $t_1$ = 2 s
This is the time in going from A to B. From B, or the point of the maximum height, the ball falls freely under the acceleration due to gravity. The ball is moving in negative y direction. We use equation
$y=y_0+v_0t+\frac{1}{2}at^2$
We have, $y_0$ = 45 m, y = 0, $v_0$ = 0, a = – g = –10 m$s^{–2}$
0 = 45 + (½) (–10)$t_2^2$
Solving, we get $t_2$ = 3 s Therefore, the total time taken by the ball before it hits the ground
= $t_1$ + $t_2$ = 2 s + 3 s = 5 s.
SECOND METHOD: The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation
$y=y_0+v_0t+\frac{1}{2}at^2$
Now $y_0$ = 25 m y = 0 m
$v_o$ = 20 m$s^{-1}$, a = –10 m$s^{–2}$, t = ?
0 = 25 + 20t + (½) (-10)$t^2$
Or, 5$t^2$ – 20t – 25 = 0
Solving this quadratic equation for t, we get t = 5s Note that the second method is better since we do not have to worry about the path of the motion as the motion is under constant acceleration.
Example 2
Stopping distance of vehicles: When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (v0) and the braking capacity, or deceleration, –a that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_o$ and a.
Answer
Let the distance travelled by the vehicle before it stops be $d_s$. Then, using equation of motion $v^2$ = $v_o^2$ + 2ax, and noting that v = 0, we have the stopping distance
$d_s=\frac{-v_0^2}{2a}$
Thus, the stopping distance is proportional to the square of the initial velocity. Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration).
For the car of a particular make, the braking distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15, 20 and 25 m/s which are nearly consistent with the above formula.
Stopping distance is an important factor considered in setting speed limits, for example, in school zones.
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