Question

A particle is moving along a straight line. Its velocity varies as v = 6 - 2t where v is in m/s and t in
seconds. If the difference between distance covered and magnitude of displacement in first 4 seconds
is 2n, then find n.
​

Answer 1

Answer:

This is the right and of this question as magnitude of displacement is distance if there is a straight line

and but they ask the difference between the distance and displacement as distance = displacement hence

2n=0

n =0

Answer 2

Answer:

The value of n is 1.

Explanation:

Given the velocity,  $v = 6 - 2t ~m/s$

Velocity is defined as rate of change of displacement.

$v=\frac{ds}{dt}\implies 6 - 2t =\frac{ds}{dt}$

$\implies ds= (6 - 2t) dt$

Integrating on both sides,

$\implies \int ds=\int\limits^4_0 (6 - 2t) dt$

$\implies s= \Big|6t - 2}\frac{t^{2}}{2} \Big|\limits^4_0$  

$= \Big|6t - t^{2} \Big|\limits^4_0$    

$=6\times 4 - 4^{2} =24-16=8m$

Displacement of the body is 8m.

And the distance covered by the body is the magnitude of velocity vector, i.e., $d=|\vec v|$

Therefore, $d=\sqrt{6^{2} -(2(4))^{2} }$

$=\sqrt{36 -64} =\sqrt{100}=10m$  

Distance covered of the body is 10m.

Given the difference between distance covered and magnitude of displacement in first 4 seconds is 2n,

$d-s=2n\implies 10-8=2n$

$\implies2n=2\implies n=1$

Hence, the value of n is 1.