Question

A particle experiences constant acceleration for 20 seconds

Answer 1

Answer:

The correct answer is D. s₂=4s₁

Explanation:

The distance of a particle is given by:

$s=s_{0}+v_{0} t + \frac{1}{2} a t^{2}$

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:

$s_{1} = s_{0}+v_{0} t_{1} + \frac{1}{2} a (t_{1})^{2}$

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

$s_{1}= \frac{1}{2} a (t_{1})^{2}$

Furthermore, the position s₂ is:

$s_{2}= \frac{1}{2} a (t_{2})^{2}$

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):

$s_{2}= \frac{1}{2} a (t_{2})^{2} \\ s_{2}= \frac{1}{2} a (2t_{1})^{2} \\ s_{2} = \frac{1}{2} a 2^{2}(t_{1})^{2} \\ s_{2} = \frac{1}{2} a 4 t_{1}^{2}$

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

$\frac{s_2}{s_1} =\frac{\frac{1}{2}a4t_{1}^{2} }{\frac{1}{2}at_{1}^{2}} =4$

$\frac{s_2}{s_1}=4 \\ s_{2} = 4 s_{1}$