Question

A particle moving in a straight line with uniform deceleration has a velocity of 40m/s at point P, 20m/s at point Q

and comes to rest at point R where QR=50m .Calculate the
1.Distance PQ
2.Time taken to cover PQ
3.Time taken to cover PR​

Answer 1

The distance between PQwill be 10,time taken to cover PQ will be 0.25sec and time taken to cover PR will be 3sec.

What is uniform deceleration?

Uniform deceleration is when velocity decrease by equal intervals of time .

Given ,

Velocity at P is 40 m/ s,

Velocity at Q is 20 m/ s

Distance between QR will be 50m.

let the whole distance between PR will be x,

so time taken of PQ, equating both times of PQ and QR we will get this,

x-50/40 = 50/ 20

after calculating ,

x = 60m , this the distance of whole straight line

60-50=. 10 m

the distance between PQ will be 10

2) time taken to cover PQ

. V= d/ t

 t= 0.25s

3) time taken to cover PR

20=60/ t

t= 3sec.

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Answer 2

1. PQ=150m.

2. PQ: $t=5s$

3. QR: $t=5s$

What is velocity?

The vector quantity velocity (v), denoted by the equation v = s/t, quantifies displacement (or change in position, s) over the change in time (t). Speed (or rate, r) is a scalar number, denoted by the equation r = d/t, that quantifies the distance traveled (d) over the change in time (t).

QR=50m.

$V_{Q} (u)=20ms^{-1} ,V_{R} v$

$v^{2} =u^{2}+2as$

$0=20^{2} +2a*50$

100a=-400

$a=-4ms^{-2}$

1. Distance PQ:

$u_{p} (u)=40msx^{-1} , v_{Q (v)=20ms^{-1}$

$v^{2} =u^{2} +2as$

$20^{2} =40^{2} -2*4s$

400=1600-400

8s=1600-400

$s=\frac{1200}{8} =150m$

PQ=150m

2.Time taken to cover PQ:

v=u+at

20=40-4t

4t=40-20

$t=\frac{20}{4}=5s$

3. Time taken to cover PR​:

v=u+at

0=20-4t

$t=\frac{20}{4}=5s$

1. PQ=150m.

2. PQ: $t=5s$

3. QR: $t=5s$

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