Question

A particle is moving with a velocity of 10m/s and having a retardation of 2m/s^2. Find displacement and distance covered in

(i)5 sec.
(ii)10 sec.
(iii)15 sec.​

Answer 1

AnsweR :

• Initial velocity (u) = 10 m/s
• Acceleration (a) = - 2 m/s²

Use 2nd equation of Kinematics,

• At t = 5s

S = ut + ½at²

S = 10*5 + ½(-2)(25)

S = 50 - 25

S = 25m

Distance = Speed * time

Distance = 10 * 5 = 50m

___________________

• At t = 10s

S = 10*10 + ½(-2)(10)²

S = 100 - 100

S = 0

Distance = 10 * 10 = 100m

__________________

• At t = 15s

S = 10*15 + ½(-2)(15)²

S = 150 - 225

S = -75 { This is not possible, as displacement can't be negative}

Distance = 10*15 = 150m

Answer 2

Answer:

Explanation:

Solution :-

1) u = 10m/s , -a = 2m/s^2

$\implies$a = -2m/s^2 ,

t = 5s

We know that :-

$s = ut + \dfrac{1}{2}\times a\times t^2$

$s = 10(5)+\dfrac{1}{2}\times (-2)\times (5)^2$

$= 50- 25$

$= 25m$

Hence , Value of Distance "s" = 25m.

'If the particle is travelling in the straight line without coming back to original position, then the displacement done is equal to distance covered'.

'Since there is no reference about motion of particle, we assume the motion of particle as a straight line' .

Therefore Value of Displacement = 25m

2) u = 10m/s , a = -2m/s^2 , t = 10s

We know that :-

$s = ut + \dfrac{1}{2}\times a\times t^2$

$= 10(10)+\dfrac{1}{2}\times (-2)\times (10)^2$

$= 100 - 100$

= 0 m

Hence , Distance('s') = 0m . Since, the body is at rest.

Then Similarly Displacement = 0m

3) u = 10m/s , a = -2m/s^2, t = 15s

We know that :-

$s = ut + \dfrac{1}{2}\times a\times t^2$

$= 10(15)+\dfrac{1}{2}\times (-2)\times (15)^2$

$= 150 - 225$

= - 75m

Here, the negative sign in the resultant distance denotes that the distance has been covered by particle in the opposite direction. Since distance can't be written in negative form, we can write its magnitude only.

$\impies |-75m| = 75m$

Therefore magnitude of the distance = 75m

'If the particle is travelling in the straight line without coming back to original position, then the displacement done is equal to distance covered'.

'Since there is no reference about motion of particle, we assume the motion of particle as a straight line' .

Since displacement is a vector quantity, It can be written in negative form.

Therefore Value of Displacement = -75m.