Question

A particle starts moving along a straight line with initial velocity of
25m/s, from O under a uniform acceleration of -2.5 m/s2. Deterime
(i) Velocity, displacement and the distance travelled at t= 5
sec
(ii) How long the particle moves in the same direction? What
is its velocity, displacement and the distance covered
then?
(iii) The instantaneous velocity , displacement and the distance
covered at t=15 sec
(iv) The time required to come back to O, velocity,
displacement and distance covered then
(v) Instantaneous velocity, , displacement and distance
covered at t=25 sec

Answer 1

Answer on Question #73991, Physics / Mechanics — Relativity —

Question A particle starts moving along a straight line with initial velocity of 25m/s, from O under a uniform acceleration of -2.5 m/s2. Deterime

(i) Velocity, displacement and the distance travelled at t= 5 sec

(ii) How long the particle moves in the same direction? What is its velocity, displacement and the distance covered then?

(iii) The instantaneous velocity , displacement and the distance covered at t=15 sec

(iv) The time required to come back to O, velocity, displacement and distance covered then

(v) Instantaneous velocity, , displacement and distance covered at t=25 sec

Solution (i)From equation for velocity:

$v(t)=v_{0}-at$

where $v_{0}=25$ m/s, $a=-2.5$ m/s^{2}. So we find:

$v(5)=25-2.5\cdot 5=12.5\,m/s$

In this case distance and displacement are the same, as particle didn’t change the direction.

$d=s(5)=v_{0}t-at^{2}/2=25\cdot 5-2.5\cdot 5^{2}/2=93.75\,m$

(ii) Time when direction is change is defined by condition:

$v(t_{x})=0$

$v_{0}-at_{x}=0$

$t_{x}=\frac{v_{0}}{a}=\frac{25}{2.5}=10\,s$

Displacement and distance:

$d=s(10)=v_{0}t-at^{2}/2=25\cdot 10-2.5\cdot 10^{2}/2=125\,m$

(iii) For $t=15$:

$v(15)=25-2.5\cdot 15=-12.5\,m/s$

The displacement is:

$s(15)=25\cdot 15-2.5\cdot 15^{2}/2=93.75\,m$

The distance is sum of displacements before and after the turn-around:

$d=s(10)+2.5\cdot 5^{2}/2=156.25\,m$

(iv) The time required to come back to O is defined by condition:

$v(t_{y})=-v_{0}$

From this we find that

$v_{0}-at_{y}=-v_{0}$

$t_{y}=\frac{2v_{0}}{a}=\frac{2\cdot 25}{2.5}=20\,s$

The displacement is obviously 0 is this case, d=0, while distance is twice as the particle has travelled to turn around:

$s(20)=2\cdot s(10)=2\cdot 125=250\,m$

(v) For $t=25$ we have:

$v(25)=25-2.5\cdot 25=-37.5\,m/s$

The distance:

$s(25)=25\cdot 25-2.5\cdot 25^{2}/2=-156.25\,m/s$

The displacement is equal to sum of displacement at 20 and displacement covered in last 5 sec:

$d=0+156.25=156.25$