Question

The displacement is s=6+12t-2t2 in 5 sec. so what is the distance

Answer 1

Answer:

The distance is 26 m.

Step-by-step explanation:

Given : The displacement is $s=6+12t-2t^2$ in 5 sec.

To find : What is the distance ?

Solution :

The displacement is $s=6+12t-2t^2$

$v=\frac{ds}{dt}$

$v=12-4t$

Equating the velocity v with zero,

$12-4t=0$

$t=3$

The particle travels for 3 seconds in forward direction and for the next 2 second in backward direction.

We have to find the magnitude of displacement in first 3 second and then 3 second to 5 second and then add the two.

Displacement in first 3s:

$S_1=s(t=3s)-s(t=0)\\S_1=(6+12\times 3-2\times 9)-6\\S_1=18m$

Displacement during 3 to 5 second:

$S_2=s(t=5s)-s(t=3s)\\S_2=(6+12\times 5-2\times25)-(6+12\times3-2\times 9)\\S_2=-8m$

Ignoring the negative sign and adding the two displacements.

$S=S_1+S_2$

$S=18+8$

$S=26$

Therefore, The distance is 26 m.