Uniformly Accelerated Motion , Non-Uniformly Accelerated Motion

Uniformly accelerated motion:

Suppose that a particle passes the origin of the coordinate system with a velocity v0 at the time t = 0 . Suppose that its displacement is s and its velocity v after time t.

Then, applying the definitions of average acceleration for constant value, average value, we get the following expressions.

$ \displaystyle (i) \vec{v} = \vec{v_0} + \vec{a} t $

$ \displaystyle (ii) \vec{S} = \vec{v_0}t + \frac{1}{2}\vec{a} t^2 $

$ \displaystyle (iii) v^2 = v_0^2 + 2\vec{a}.\vec{S} $

$ \displaystyle (iv) \vec{S_n} = \vec{v_0} + \frac{\vec{a}}{2} (2 n -1) $

Solved Example : A particle moves in a straight line with constant acceleration. If it covers 10 m in first second and extra 10 m in next second, find its initial velocity.

(a) What is the acceleration of the particle ?

(b) What is the velocity of the particle at the end of 3rd second ?

(c) What is the displacement of the particle in 3 seconds ?

(d) What is the displacement of the particle in 3rd second ?

Solution: Let the initial speed be v0 and the acceleration be a

$\large s = v_0 t + \frac{1}{2}a t^2 $

Putting t = 1 sec. and s = 10 m, we obtain

10 = v0 + 5 a …(1)

Again in next second, that means t = (1+1) sec [t is the time measured from the initial position (when t = 0, s = 0)] the displacement of the particles = 10 + 10 = 20 m.

Putting t = 2 sec and s = 20 m, we obtain

10 = 2 v0 + 2 a

v0 + a = 5 …(2)

Solving (1) and (2), we obtain

10 = v0 + 5 (5 – v0)

10 = -4 v0 + 25

v0 = 15/4 = 3.75 m/s

(a) a = 5 – v0

= 5 – 3.75 = 1.25 m/sec2

(b) v = v0 + a t

= 3.75 + (1.25)(3) = 7.5 m/sec.

(c) $\large s = v_0 t + \frac{1}{2}a t^2 $

$\large s = 3.75 \times 3 + \frac{1}{2}\times 1.25 \times 3^2 $

(d) $\large s = v_0 + \frac{a}{2}(2 t – 1)$

$\large s = 3.75 + \frac{1.25}{2}(2 \times 3 – 1)$

Non-uniformly accelerated motion in a straight line:

If the acceleration a is a function of time (t) or function of distance (s) the equation of instantaneous velocity and acceleration can be written as

If , a = f(t)

$ \displaystyle v = \frac{ds}{dt} $

$ \displaystyle s = \int_{0}^{t} vdt $

$ \displaystyle a = \frac{dv}{dt} $

$ \displaystyle v = \int_{0}^{t} a dt + v_0 $

If , a = f(s)

$ \displaystyle a = \frac{dv}{dt} = \frac{dv}{ds}.\frac{ds}{dt} $

(considering the magnitudes we obtain a scalar equation.)

$ \displaystyle a = v \frac{dv}{ds} \quad (Since ,\frac{ds}{dt} = v ) $

$ \displaystyle \int_{v_0}^{v} v dv = \int_{0}^{S} a ds $

$ \displaystyle \frac{v^2 – v_0^2}{2} = \int_{0}^{S} a ds $

$ \displaystyle v^2 = v_0^2 + 2 \int_{0}^{S} a ds $

Example : The acceleration of a particle moving rectilinearly varies with the magnitude of its velocity as a = – √v . Find its initial speed , if it stops after to = 1 sec from starting .

(i) Find the velocity-time equation.

(ii) Find the displacement-time equation.

Solution : $ a = – \sqrt{v} $

$\large \frac{dv}{dt} = – \sqrt{v} $

$\large \frac{dv}{\sqrt{v}} = –  dt$

$\large \int_{v_0}^{v} \frac{dv}{\sqrt{v}} = –  \int_{0}^{t} dt$

$\large 2 (\sqrt{v} – \sqrt{v_0}) = – t $

$\large 2 (\sqrt{v_0} – \sqrt{v}) =  t $

At   t = t0, it stops; Putting v = 0,  we obtain

$\large t_0 = 2\sqrt{v_0} $

$\large v_0 = \frac{t_0^2}{2} $

putting t0= 1, we obtain v0 = 0.5 m/sec.

(i) $ a = – \sqrt{v} $

$\large \frac{dv}{dt} = – \sqrt{v} $

$\large \frac{dv}{\sqrt{v}} = –  dt$

$\large \int_{v_0}^{v} \frac{dv}{\sqrt{v}} = –  \int_{0}^{t} dt$

$\large 2 (\sqrt{v} – \sqrt{v_0}) = – t $

$\large 2 (\sqrt{v_0} – \sqrt{v}) =  t $

$\large \sqrt{v} = \sqrt{v_0} – \frac{t}{2} $

$\large v = v_0 + \frac{t^2}{4} – \sqrt{v_0} t $

(ii) $\large v = \frac{dx}{dt} $

$\large x = \int_{0}^{t} v dt $

Now , put the obtained expression of v

Also Read :

→ Rest & Motion Relative to Frame of Reference
→ Average Speed & Average Velocity
→ Average Acceleration , Instantaneous Acceleration , Average Speed & Average Velocity
→ Vertical motion under gravity
→ Relative Velocity

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