3.6 Kinematic equations for uniformly accelerated motion

KINEMATIC EQUATIONS FOR UNIFORMLY ACCELERATED MOTION(MOTION IN A STRAIGHT LINE)

For uniformly accelerated motion, an object moving with uniform acceleration \(a\): Let the velocity at time 0 be \(v_{0}\) and the velocity at time \(t\) be \(v\). Thus,
\(
\begin{aligned}
\frac{d v}{d t} &=a, \quad \text { or, } \quad d v=a d t \\
\text { or, } \quad & \int_{v_{0}}^{v} d v =\int_{0}^{t} a d t .
\end{aligned}
\)

As time changes from 0 to \(t\) the velocity changes from \(v_{0}\) to \(v\). Evaluating the integrals we get,
\(
\begin{aligned}
& {[v]_{v_{0}}^{v} } &=a[t]_{0}^{t} \\
\text { or, } & v-v_{0} =a t
\end{aligned}
\)
\(
\text { or, } \quad v=v_{0}+a t .
\)

This relationship is graphically represented in the figure below.

Figure 3.1: The area under the v-t curve for an object with uniform acceleration

The area under the above curve is :
The area between instants 0 and t = Area of triangle ABC + Area of rectangle OACD

The area under \(v\) – \(t\) curve represents the displacement. Therefore, the displacement \(x\) of the object is:
\(x=\frac{1}{2}\left(v-v_{0}\right) t+v_{0} t\)
But \(v-v_{0}=a t\)
Therefore, \(x=\frac{1}{2} a t^{2}+v_{0} t\)
or, \(x=v_{0} t+\frac{1}{2} a t^{2}\)

Squaring both sides of the equation \(v=v_{0}+a t \), we get

\(\begin{aligned}
v^{2} &=(v_{0}+a t)^{2} \\
&=v_{0}^{2}+2 v_{0} a t+a^{2} t^{2} \\
&=v_{0}^{2}+2 a\left(v_{0} t+\frac{1}{2} a t^{2}\right) \\
&=v_{0}^{2}+2 a x .
\end{aligned}\)

Points To Remember: The three kinematic equations are very useful in solving the problems of motion in a straight line with constant acceleration.

\(\begin{aligned}
&v=v_{0}+a t \\
&x=v_{0} t+\frac{1}{2} a t^{2} \\
&v^{2}=v_{0}^{2}+2 a x
\end{aligned}\)

The above kinematic equations 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 above equations are modified (replacing \(x\) by \(x-x_{0}\) ) to:
\(
\begin{aligned}
&v=v_{0}+a t \\
&x=x_{0}+v_{0} t+\frac{1}{2} a t^{2} \\
&v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right)
\end{aligned}
\)

Example 3.4:

A ball is thrown vertically upwards with a velocity of \(20 \mathrm{~m} \mathrm{~s}^{-1}\) from the top of a multistorey building. The height of the point from where the ball is thrown is \(25.0 \mathrm{~m}\) from the ground. (a) How high will the ball rise? and (b) how long will it be before the ball hits the ground? Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).

Answer: (a) Let us take the \(y\)-axis in the vertically upward direction with zero at the ground, as shown in Fig. 3.2 below.

Now \(v_{o}=+20 \mathrm{~m} \mathrm{~s}^{-1}\),
\(
a=-g=-10 \mathrm{~m} \mathrm{~s}^{-2},
\)
\(
v=0 \mathrm{~m} \mathrm{~s}^{-1}
\)
If the ball rises to height \(y\) from the point of launch, then using the equation
\(
v^{2}=v_{0}^{2}+2 a\left(y-y_{0}\right)
\)
we get
\(
0=(20)^{2}+2(-10)\left(y-y_{0}\right)
\)
Solving, we get, \(\left(y-y_{0}\right)=20 \mathrm{~m}\).

(b) let’s split the path into 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:
\(
\begin{gathered}
v=v_{\mathrm{o}}+a t \\
0=20-10 t_{1} \\
t_{1}=2 \mathrm{~s}
\end{gathered}
\)
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 the negative \(y\) direction. We use equation
\(
y=y_{0}+v_{0} t+\frac{1}{2} a t^{2}
\)
We have, \(y_{0}=45 \mathrm{~m}, y=0, v_{0}=0, a=-g=-10 \mathrm{~m} \mathrm{~s}^{-2}\)
\(
0=45+(1 / 2)(-10) t_{2}^{2}
\)
Solving, we get \(t_{2}=3 \mathrm{~s}\)
Therefore, the total time taken by the ball before it hits the ground \(=t_{1}+t_{2}=2 \mathrm{~s}+3 \mathrm{~s}=5 \mathrm{~s}\).

FREELY FALLING BODIES

A common example of motion in a straight line with constant acceleration is the free fall of a body near the earth’s surface. If air resistance is neglected and a body is dropped near the surface of the earth, it falls along a vertical straight line. The acceleration is in the vertically downward direction and its magnitude is almost constant if the height is small as compared with the radius of the earth \(\left(6400 \mathrm{~km}\right)\). This magnitude is approximately equal to \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) or \(32 \mathrm{ft} / \mathrm{s}^{2}\) and is denoted by the letter \(g\).

If we take vertically upward as the positive \(Y\)-axis, acceleration is along the negative \(Y\)-axis and we write \(a=-g\). The kinematics equations can be written as
\(
\begin{aligned}
v &=v_{0}-g t \\
y &=v_{0} t-\frac{1}{2} g t^{2} \\
v^{2} &=v_{0}^{2}-2 g y .
\end{aligned}
\)
Here \(y\) is the \(y\)-coordinate (that is the height above the origin) at time \(t, v_{0}\) is the velocity in \(y\) direction at \(t=0\) and \(v\) is the velocity in \(y\) direction at time \(t\). The position of the particle at \(t=0\) is \(y=0\).

Example 3.5:

A ball is thrown up at a speed of \(4.0 \mathrm{~m} / \mathrm{s}\). Find the maximum height reached by the ball. Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\).

Answer: Let us take vertically upward direction as the positive \(Y\)-axis. We have \(u=4.0 \mathrm{~m} / \mathrm{s}\) and \(a=-10 \mathrm{~m} / \mathrm{s}^{2}\). At the highest point the velocity becomes zero. Using the kinematics formula.

\(\begin{aligned}
&v^{2}=v_{0}^{2}+2 a y \\
&0=(4 \cdot 0 \mathrm{~m} / \mathrm{s})^{2}+2\left(-10 \mathrm{~m} / \mathrm{s}^{2}\right) y \\
&y=\frac{16 \mathrm{~m}{ }^{2} / \mathrm{s}^{2}}{20 \mathrm{~m} / \mathrm{s}^{2}}=0.80 \mathrm{~m}
\end{aligned}\)

Example 3.6: Free-fall: Discuss the motion of an object under free fall. Neglect air resistance.

Answer: An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of the acceleration due to gravity is represented by g. If air resistance is neglected, the object is said to be in free fall. If the height through which the object falls is small compared to the earth’s radius, \(g\) can be taken to be constant, equal to \(9.8 \mathrm{~m} \mathrm{~s}^{-2}\). Free fall is thus a case of motion with uniform acceleration.
We assume that the motion is in \(y\)-direction, more correctly in – \(y\)-direction because we choose upward direction as positive. Since the acceleration due to gravity is always downward, it is in the negative direction and we have \(a=-g=-9.8 \mathrm{~m} \mathrm{~s}^{-2}\)
The object is released from rest at \(y=0\). Therefore, \(v_0=0\) and the equations of motion become:
\(
\begin{array}{lll}
v=0-g t & =-9.8 t & \mathrm{~m} \mathrm{~s}^{-1} \\
y=0-1 / 2 g t^2 & =-4.9 t^2 & \mathrm{~m} \\
v^2=0-2 g y & =-19.6 y \quad \mathrm{~m}^2 \mathrm{~s}^{-2}
\end{array}
\)
These equations give the velocity and the distance travelled as a function of time and also the variation of velocity with distance. The variation of acceleration, velocity, and distance, with time have been plotted in Fig. 3.14(a), (b) and (c).

Galileo’s law of odd numbers

“The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7 ………]. Let us prove this.

Let us divide the time interval of motion of an object under free fall into many equal intervals \(\tau\) and find out the distances traversed during successive intervals of time. Since the initial velocity is zero, we have
\(
y=-\frac{1}{2} g t^2
\)
Using this equation, we can calculate the position of the object after different time intervals, \(0, \tau, 2 \tau, 3 \tau \ldots\) which are given in the second column of Table 3.2. If we take \((-1 / 2) g \tau^2\) as \(y_0\) – the position coordinate after first-time interval \(\tau\), then the third column gives the positions in the unit of \(y_{o}\). The fourth column gives the distances traversed in successive \(\tau\) s. We find that the distances are in the simple ratio \(1: 3: 5: 7: 9: 11 \ldots\) as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall.

Example 3.7: 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 \(\left(v_0\right)\) 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 \(v_0\) and \(a\).

Answer: Let the distance travelled by the vehicle before it stops be \(d_{s}\). Then, using equation of motion \(v^2=v_0^2+2 a x\), and noting that \(v=0\), we have the stopping distance
\(
d_s=\frac{-v_0^2}{2 a}
\)
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 \(10 \mathrm{~m}, 20 \mathrm{~m}, 34 \mathrm{~m}\) and \(50 \mathrm{~m}\) corresponding to velocities of \(11,15,20\) and \(25 \mathrm{~m} / \mathrm{s}\) which are nearly consistent with the above formula.

Example 3.8: Reaction time: When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on the complexity of the situation and on the individual.

You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger (Fig. below). After you catch it, find the distance \(d\) travelled by the ruler. In a particular case, \(d\) was found to be \(21.0 \mathrm{~cm}\). Estimate reaction time.

Answer: The ruler drops under free fall. Therefore, \(v_o=0\), and \(a=-g=-9.8 \mathrm{~m} \mathrm{~s}^{-2}\). The distance travelled \(d\) and the reaction time \(t_r\) are related by \(
d=-\frac{1}{2} g t_r^2
\)
Or, \(\quad t_r=\sqrt{\frac{2 d}{g}} \mathrm{~s}\)
Given \(d=21.0 \mathrm{~cm}\) and \(g=9.8 \mathrm{~m} \mathrm{~s}^{-2}\) the reaction time is
\(
t_r=\sqrt{\frac{2 \times 0.21}{9.8}} \mathrm{~s} \cong 0.2 \mathrm{~s} .
\)

Example 3.9: A particle starts with an initial velocity \(25 \mathrm{~m} / \mathrm{s}\) along the positive \(x\) direction and it accelerates uniformly at the rate \(0.50 \mathrm{~m} / \mathrm{s}^2\). (a) Find the distance travelled by it in the first two seconds. (b) How much time does it take to reach the velocity \(7.5 \mathrm{~m} / \mathrm{s}\)? (c) How much distance will it cover in reaching the velocity \(7.5 \mathrm{~m} / \mathrm{s}\)?

Answer: (a) We have,
\(
\begin{aligned}
x & =u t+\frac{1}{2} a t^2 \\
& =(2.5 \mathrm{~m} / \mathrm{s})(2 \mathrm{~s})+\frac{1}{2}\left(0.50 \mathrm{~m} / \mathrm{s}^2\right)(2 \mathrm{~s})^2 \\
& =5.0 \mathrm{~m}+1.0 \mathrm{~m}=6.0 \mathrm{~m} .
\end{aligned}
\)
Since the particle does not turn back it is also the distance travelled.
(b) We have,
\(
\begin{gathered}
v=u+a t \\
\text { or, } \quad 7.5 \mathrm{~m} / \mathrm{s}=2.5 \mathrm{~m} / \mathrm{s}+\left(0.50 \mathrm{~m} / \mathrm{s}^2\right) t \\
\text { or, } \quad t=\frac{7.5 \mathrm{~m} / \mathrm{s}-2.5 \mathrm{~m} / \mathrm{s}}{0.50 \mathrm{~m} / \mathrm{s}^2}=10 \mathrm{~s}
\end{gathered}
\)
(c) We have,
\(
\begin{array}{rlrl}
v^2 & =u^2+2 a x \\
\text { or, } \quad(7.5 \mathrm{~m} / \mathrm{s})^2 & =(2.5 \mathrm{~m} / \mathrm{s})^2+2\left(0.50 \mathrm{~m} / \mathrm{s}^2\right) x \\
\text { or, } & x =\frac{(7.5 \mathrm{~m} / \mathrm{s})^2-(2.5 \mathrm{~m} / \mathrm{s})^2}{2 \times 0.50 \mathrm{~m} / \mathrm{s}^2}=50 \mathrm{~m} .
\end{array}
\)

Example 3.10: A particle having initial velocity \(u\) moves with a constant acceleration a for a time \(t\). (a) Find the displacement of the particle in the last 1 second. (b) Evaluate it for \(u=5 \mathrm{~m} / \mathrm{s}, a=2 \mathrm{~m} / \mathrm{s}^2\) and \(t=10 \mathrm{~s}\).

Answer: (a) The position at time \(t\) is
\(
s=u t+\frac{1}{2} a t^2
\)
The position at time \((t-1 \mathrm{~s})\) is
\(
\begin{aligned}
s^{\prime} & =u(t-1 \mathrm{~s})+\frac{1}{2} a(t-1 \mathrm{~s})^2 \\
& =u t-u(1 \mathrm{~s})+\frac{1}{2} a t^2-a t(1 \mathrm{~s})+\frac{1}{2} a(1 \mathrm{~s})^2
\end{aligned}
\)
Thus, the displacement in the last \(1 \mathrm{~s}\) is
\(
\begin{aligned}
s_t & =s-s^{\prime} \\
& =u(1 \mathrm{~s})+a t(1 \mathrm{~s})-\frac{1}{2} a(1 \mathrm{~s})^2 \\
\text { or, } \quad s_t & =u(1 \mathrm{~s})+\frac{a}{2}(2 t-1 \mathrm{~s})(1 \mathrm{~s}) .
\end{aligned}
\)
(b) Putting the given values in (i)
\(
\begin{aligned}
s_t & =\left(5 \frac{\mathrm{m}}{\mathrm{s}}\right)(1 \mathrm{~s})+\frac{1}{2}\left(2 \frac{\mathrm{m}}{\mathrm{s}^2}\right)(2 \times 10 \mathrm{~s}-1 \mathrm{~s})(1 \mathrm{~s}) \\
& =5 \mathrm{~m}+\left(1 \frac{\mathrm{m}}{\mathrm{s}^2}\right)(19 \mathrm{~s})(1 \mathrm{~s}) \\
& =5 \mathrm{~m}+19 \mathrm{~m}=24 \mathrm{~m} .
\end{aligned}
\)