Motion in Straight Line

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For uniformly accelerated motion, derivation related to displacement (x), the time taken (t), initial velocity (v0), final velocity (v) and acceleration (a).

The average acceleration equals the constant value of acceleration during the interval. If the
velocity of an object is v_o at t = 0 and v  at time t,

we have,

$$\begin{gathered} \overline{a} = \frac{v-v_o}{t- 0} \\ v = v_o +at \end{gathered}$$ ...(i)

This relation is graphically represented in Figure

The area under this curve is :

Area between instants 0 and t = Area of triangle ABC + Area of rectangle OACD

Area under the v-t curve for an object with uniform acceleration

The area under v-t curve represents the displacement.

Therefore, the displacement x of the object is :

$$\begin{gathered} x = \frac{1}{2}(v-v_o)t +v_{o}t ........\left(ii\right) \\ But, v-v_o = at \\ \therefore x = \frac{1}{2} a{t}^2 + v_{0}t \\ or, x = v_{0}t + \frac{1}{2}a{t}^2 ..\left(iii\right) \\ Equ. \left(ii\right) can be written as, \\ x = \frac{v+v_0}{2}t =\overline{v}t .... \left(iv\right) \\ where, \\ \overline{v} = \frac{v +v_o}{2} ........\left(v\right) \end{gathered}$$

Equations (iv) and (v) mean that the object has undergone displacement x with an average velocity equal to the arithmetic average of the initial and final velocities.

From equation (i)  t = (v – v₀)/a. Substituting this in Eq. (iv), we get

$$\begin{gathered} x = \overline{v}t \\ = \frac{v+ v_o}{2} \frac{v-v_0}{a} = \frac{v^2-v_0^2}{2a} \\ {v}^2 = v_0^2 + 2ax ....\left(vi\right) \end{gathered}$$

This equation can also be obtained by substituting the value of t from Eq. (i) into Eq. (iii). Thus, we have obtained three important equations :

$$\begin{gathered} v = v_o +at \\ x = v_o t + \frac{1}{2}a{t}^2 \\ {v}^2 = v_0^2 + 2ax \end{gathered}$$

Then Eqs. (vi) are modified (replacing x by x – x₀) to :

$v^2 = v_0^2 +2a(x-x_o)$