Question

25. A particle experiences a constant acceleration for 6 seconds after starting from rest. If it travels a distance
d1in the first two seconds, d2 in the next two seconds and a distance d3in the last 2 seconds, then
(1) d1:d₂:d₃= 1 : 1:2
(2) d1 : d₂:d₃=1:2:3
(3) d1: d2: d3 = 1:3:5
(4) d1 : d₂:d3 = 1:5:9​

Answer 1

Answer:

number 3

Explanation:

Answer 2

Answer:(3)

Explanation:

Let velocity after 2 second is V1

Hence V1=u+at where a is acceleration which is constant..

After 2 second V1=2×a. (Particle is initially at rest)

Again we know v^2=u^2+2ad hence d1 =v1^2/2*a

d1=(2a)^2/2a

d1=2a

After 2 second velocity is 2a.(it take as initial velocity for the second case).let velocity after next 2 second is V2

Hnece V2=V1+at

V2=2a+a*2

V2=4a

Again v2^2=v1^2+2ad2

d2=(v2^2-v1^2)/2*a

d2=(4a)^2-(2a)^2/2*a

d2=6a

For the third case initial velocity V2

Hence V3=V2+at

V3=4a+2a

V3=6a

And (v3)^2=(V2)^2+2ad3

d3=(v3)^2-(V2)^2/2*a

d3=(6a)^2-(4a)^2/2*a

d3=10a

Hence d1:d2:d3=2a:6a:10a

d1:d2:d3=1:3:5