Question

If A is the magnitude of the displacement covered by the particle from $t=4s$ to $t=7s$, B is the magnitude of the distance covered from from $t=4s$ to $t=7s$ and C is the magnitude of the final position of the particle w.r.t the origin, then A : B : C will be

• A. 1 : 1 : 3
• B. 5 : 5 : 16
• C. 3 : 5 : 16
• D. 3 : 5 : 15

Answer 1

The correct option is D 3 : 5 : 15



Area under the velocity-time graph is displacement.

So magnitude of displacement from $4-7s$

$A=$ Area under the graph from $4-6s$ + Area under the graph from $6-7s$

$=\frac{1}{2}\times 2\times 4-\frac{1}{2}\times 2\times 1=3m$

Distance covered from $4-7s$

$B=$ Area under the graph from $4-6s$ + Magnitude of area under the graph from $6-7s$

$=\frac{1}{2}\times 2\times 4-(-\frac{1}{2}\times 2\times 1)=5m$



Magnitude of the final position w.r.t origin
$C$= Total area under the graph $=\frac{1}{2}\times 4\times 2+4\times 2+\frac{1}{2}\times 4\times 2-\frac{1}{2}\times 2\times 1$

(because motion is backward from $6-7s$)

$=16-1=15m$

So, $A:B:C=3:5:15$