Question

A particle experiences constant acceleration for 20 seconds

after starting from rest. If it travels a distance s1 in the first

10 seconds and distance s2 in the next 10 seconds then

a. s2 = s1

b. s2 = 2s1

c. s2 = 3s1

d. s2 = 4s1

Why?​

Answer 1

The correct answer is D. s₂=4s₁

Step-by-step explanation:

The distance of a particle is given by:

`s=s_(0)+v_(0) t + (1)/(2) a t^(2)`

where

s₀ is the initial position when t=0

v₀ is the initial speed when t=0

a is the constant acceleration

t is the time in seconds

Then, the position s₁ is given by:

`s_(1) = s_(0)+v_(0) t_(1) + (1)/(2) a (t_(1))^(2)`

As the particle starts from rest v₀=0 and we consider s₀=0, s₁ will be:

`s_(1)= (1)/(2) a (t_(1))^(2)`

Furthermore, the position s₂ is:

`s_(2)= (1)/(2) a (t_(2))^(2)`

In this case t₁=10 s and t₂=20 s (10 seconds later). In other words t₂=2t₁.

We replace the value of t₂ in the second equation (s₂):

`s_(2)= (1)/(2) a (t_(2))^(2) \\ s_(2)= (1)/(2) a (2t_(1))^(2) \\ s_(2) = (1)/(2) a 2^(2)(t_(1))^(2) \\ s_(2) = (1)/(2) a 4 t_(1)^(2)`

Finally, we divide s₂ by s₁ to get the ratio:

`(s_2)/(s_1) =((1)/(2)a4t_(1)^(2) )/((1)/(2)at_(1)^(2)) =4`

`(s_2)/(s_1)=4 \\  s_(2) = 4 s_(1)`