What is v-t graph? Derive the expression `x = V_0 t + 1//2 at^2` using v-t graph.

Explanation

A graph of velocity of a particle plotted along the Y-axis and time along the X-axis is known as v-t graph and the curve is known as velocity time curve.

Let `'v_theta'` be the initial velocity of a particle. Let 'a' be the uniform acceleration `v-t` graph given a straight line with a constant slope, `tan theta=m`
i.e. `a=(v-v_theta)/(t)............(1)`
From the figure `OABD` is a trapezium.
Area of trapezium OABD = Area of rectangle OACD + Area of triangle ACB
The second term on the right indicates the addition distance covered by the particle due to acceleration.
i.e. area of trapezium `=(OA)(OD)+1//2(AC)(BC)=v__(0)t+1//2t(v-v_(0))` by using (1)
Are of trapezium `=v_(0)t+1//2 at^2`
From the dimensional analysis, the right hand terms indicate the distance travelled. From the principle of homogeneity, the left hand side term should indicate the distance covered.
Hence area of the trapezium `=x=v_(0)t+1//2 at^2`
For a uniform motion, `a=0 and x=v_(0) t`
For a particle starting from rest `v_0=0, x=1//2 at^2`
In a vector form, `overset(to)(x)=overset(to)(v_0)t+(1)/(2) overset(to)(a)t^2`
Note : For a verticle upward motion, `y=v_(0)t-(1)/(2) gt^2` and for a verticle downward motion and `-y=-v_(0)t-(1)/(2) "gt"^2.` i.e., `y=v_(0)t+(1)/(2) "gt"^2`.