IIT JEE , AIEEE Forum - Physics , Mechanics

studen9t_iit

The position of a particle along the x axis at atime t is given by x=2+t-3(t^2). The displacement and distance travelled in the interval t=0 to t=1 are respectively :

a) 2 , 2

b) -2 , 2.5

c) 2 , 0

d) -2 , 2.16

SCIENTIST135

AS IT TRAVELS IN A STRAIT LINE, BOTH ARE SAME. WE CAN TAKE DISPLACEMENT AS POSITIVE BECAUSE, INITIAL DIRECTION WAS NOT SPECIFIED. ANS I S A OPTION. JUST PUT t=0 and t=1. you will get final and initial positions.

karthik2007

For finding out the displacement, simply put t=1 and t=0, and subtract the two quantities,

To find the distance, do this:

We are given : x(t) = 2+t-3t^2

Now, , ie, at this time, the displacement changes its sign.

So, the distance can now be found out as $d = |\int^{\frac{1}{6}}_0 2+t-3t^2 \, dt| + |\int^1_{\frac{1}{6}} 2+t-3t^2 \, dt|$

(Take mod of both the integrals)

sahilmadaan12

karthik is absolutely right

mukulaish

yup karthik is correct.. but the answer is not cumin that way.....

studen9t_iit

thanks..i got my answer its very similar to what karthik has done..

Ask & Discuss Questions with Community & Experts