Question

A particle moving in a straight line such that its velocity
is given by v = 6 - 3t where v is in ms 1 and t is in
second. The ratio of distance to the displacement of
the particle in first 3s is​

Answer 1

Explanation:

The relation is

d t

d x

=

v

where x is displacement and v is velocity which is given as

v=

3−

t

so we have

dx=

vdt

after 3sec the direction of motion will get reversed.

so

on integrating both sides we get

∫

X=0

X

dx=

∫

t=0

t=3

(3−

t)dt=

9−

2

9

=

4.5meter

now integrating both sides for

t=

3

to

t=

5

we have the displacement as

∫

X=4.5

X

dx= ∫

t=3

t=5

(3− t)dt= −2meter

so the net distance will be

X=

(2)+

4.5=

6.5meter