3.3 Motion in a straight line with non-uniform acceleration

  Motion in One Dimension

    1.0 Introduction
    2.0 Kinematic variables
    3.0 Motion in one dimension
    4.0 Derivation of the kinematics equation
    5.0 Vertical motion under gravity
    6.0 Analysis of motion through graph
    7.0 Relative motion
    8.0 Simultaneous motion of two bodies
    9.0 River boat problem
    10.0 Aircraft-wind problem
    11.0 Rain problem

---

3.3 Motion in a straight line with non-uniform acceleration

When the acceleration is not uniform, we cannot use the basics kinematics equations.

So, in case of motion with non-uniform acceleration, we use the following equations.
$$\overrightarrow v = \frac{{d\overrightarrow s }}{{dt}}\quad {\text{or}}\quad \int {d\overrightarrow s } = \int {\overrightarrow v dt} $$$$\overrightarrow a = \frac{{d\overrightarrow v }}{{dt}}\quad {\text{or}}\quad \int {d\overrightarrow v } = \int {\overrightarrow a dt} $$
Note:

• The above equations are valid for all types of motion whether it is accelerated or non-accelerated.

Also, it is easy to remember the relation between displacement, velocity, and acceleration as given below.

---

Question: The displacement-time equation of a particle moving along $x-$ axis is $x = {t^2} - 5t + 45$ where $x$ is in metres and $t$ is in seconds. Find the,

(a). velocity of the particle at any time $t$
(b). velocity of the particle at $t=2$
(c). time at which the particle is at rest
(d). acceleration of the particle at any time $t$
(e) acceleration at $t=2s$

Solution: The displacement equation of particle is given as,
$$x = {t^2} - 5t + 45$$
(a). We know velocity can be obtained by differentiating the displacement equation.
$$\frac{{dx}}{{dt}} = 2t - 5$$
So velocity at any time $t$ is,$$v = 2t - 5$$
(b). Velocity of the particle at $t=2s$, $$v = 2\left( 2 \right) - 5$$$$v = - 1m/s$$
(c). When the velocity of the particle is zero, it comes to rest. So, the time at which the particle is at rest is,
$$0 = 2t - 5$$$$t = 2.5s$$
(d). We know acceleration can be obtained by differentiating the velocity equation.
$$\frac{{dv}}{{dt}} = 2$$
So acceleration at any time $t$ is, $$a=2$$
(e). Since acceleration does not vary with time. It is constant throughout the motion.

Therefore, acceleration at $t=2s$ is,
$$a=2\ m/s^2$$

---

Question: A particle at rest starts moving along positive $x-$ axis with an acceleration $a=6t$.

(a). Find the velocity of the particle at any time $t$
(b). Find the position of the particle at any time $t$ if the particle was initially at $x=3$.

Solution:

At time $t=0,\ v=0,\ x=3$ (boundary conditions)

(a). We know acceleration of the particle is,
$$a=3t$$As, $a=\frac{dv}{dt}$$$\frac{{dv}}{{dt}} = 6t$$$$dv = 6tdt$$ntegrating both sides we get,$$\int {dv} = 6\int {tdt} $$$$v = 6\left[ {\frac{{{t^2}}}{2}} \right] + c$$or$$v = 3{t^2} + c$$ We will find the value of $c$ by using boundary conditions. We know that the particle is at rest $(v=0)$ at time $t=0$.
$$0 = 0 + c$$$$c = 0$$ So, the final velocity equation of the particle is, $$v = 3{t^2} \quad ...(i)$$

(b). Relation between position and velocity of the particle is, $$\frac{{dx}}{{dt}} = v \quad ...(ii)$$
So, from equation $(i)$ and $(ii)$ we get, $$\frac{{dx}}{{dt}} = 3{t^2}$$$$dx = 3{t^2}dt$$$$x = {t^3} + k$$ We will find the value of $k$ by using boundary condition. We know that the particle is at $x=3$ at time $t=0$.

So, $$3 = 0 + k$$$$k = 3$$ Therefore, the equation becomes,
$$x = {t^3} + 3$$

---

Question: A train starting from rest is accelerated and the instantaneous acceleration is given by $\frac{{10}}{{v + 1}}m/{s^2}$, where $v$ is the velocity in $m/s$.

(a). Find the time after which the body attains a velocity of $54\ km/hr$.
(b). Find the distance in which the train attains a velocity of $54\ km/hr$.

Solution: We will first convert $km/hr$ in $m/s$.
$$54km/hr = 54 \times \frac{{1000}}{{3600}} = 15m/s$$
Acceleration of the particle is given as, $$a = \frac{{10}}{{v + 1}} \quad ...(i)$$As $\left( {a = \frac{{dv}}{{dt}}} \right)$ $$\frac{{dv}}{{dt}} = \frac{{10}}{{v + 1}}$$$$\left( {v + 1} \right)dv = 10dt$$Integrating both sides we get, $$\int {\left( {v + 1} \right)dv} = 10\int {dt} $$$$\frac{{{v^2}}}{2} + v = 10t + c$$

Particle is at rest $(v-0)$ at time $t=0$. So, $$0 + 0 = 0 + c$$$$c = 0$$ So, the final velocity equation is, $$\frac{{{v^2}}}{2} + v = 10t$$

(a). Time taken by the body to attain a velocity of $15\ m/s$ or $54\ km/hr$.
$$10t = \frac{{{{\left( {15} \right)}^2}}}{2} + 15$$$$10t = 112.5 + 15$$$$10t = 127.5$$$$t = 12.75s$$
(b). The acceleration of the particle is given as, $$a = \frac{{dv}}{{dt}}$$or$$a = \frac{{dv}}{{ds}} \times \frac{{ds}}{{dt}}$$$$a = \frac{{dv}}{{ds}}v$$or$$a = v\frac{{dv}}{{ds}} \quad ...(ii)$$From equation $(i)$ and $(ii)$ we get, $$\frac{{10}}{{v + 1}} = v\frac{{dv}}{{ds}}$$$$10ds = v\left( {v + 1} \right)dv$$$$10ds = \left( {{v^2} + v} \right)dv$$Integrating both sides we get, $$10\int {ds} = \int {\left( {{v^2} + v} \right)dv} $$$$10s = \frac{{{v^2}}}{3} + \frac{{{v^2}}}{2} + c$$At $v=0$, $s=0$. So, $$0 = 0 + 0 + c$$$$c = 0$$Therefore the equation becomes, $$s = \frac{1}{{10}}\left( {\frac{{{v^3}}}{3} + \frac{{{v^2}}}{2}} \right)$$At $v=15\ m/s$ displacement is, $$s = \frac{1}{{10}}\left( {\frac{{{{15}^3}}}{3} + \frac{{{{15}^2}}}{2}} \right)$$$$s = \frac{1}{{10}}\left( {\frac{{3375}}{3} + \frac{{225}}{2}} \right)$$$$s = \frac{1}{{10}}\left( {1125 + 112.5} \right)$$$$s = \frac{1}{{10}} \times 1237.5$$$$s = 123.75m$$