Question

What does the area of an 'acceleration-displacement' graph represent?

• A. Distance
• B. Velocity
• C. $\frac{v^2-u^2}{2}$
• D. $\frac{v-u}{t}$

Answer 1

The correct option is C $\frac{v^2-u^2}{2}$

$a=dv/dt.$

Multiply and divide RHS by dx and with little rearrangement we get,

$a=\left(dx/dt\right)x\left(dv/dx\right)=v\left(dv/dx\right)$

Cross multuplying will give,

$integration(a.dx)=integration(v.dv)$

Now,

LHS equates to area under acceleration-distance curve, which in this case is a trapezium whose area is given by $A=\left(1/2\right)x(sum.of.parallel.sides)xheigh$

while

RHS work outs to $\left(v^2\right)/2$ which reduces to $\left[\frac{(v^2-u^2)}{2}\right]$ where v is final velocity and u is initial velocity.