Answer:
The correct answer is D. s₂=4s₁
Explanation:
The distance of a particle is given by:
[tex]s=s_{0}+v_{0} t + \frac{1}{2} a t^{2}[/tex]
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:
[tex] s_{1} = s_{0}+v_{0} t_{1} + \frac{1}{2} a (t_{1})^{2} [/tex]
As the particle starts from rest v₀=0 and we consider s₀=0, s₁ will be:
[tex] s_{1}= \frac{1}{2} a (t_{1})^{2} [/tex]
Furthermore, the position s₂ is:
[tex] s_{2}= \frac{1}{2} a (t_{2})^{2} [/tex]
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):
[tex] 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} [/tex]
Finally, we divide s₂ by s₁ to get the ratio:
[tex] \frac{s_2}{s_1} =\frac{\frac{1}{2}a4t_{1}^{2} }{\frac{1}{2}at_{1}^{2}} =4 [/tex]
[tex]\frac{s_2}{s_1}=4 \\ s_{2} = 4 s_{1}[/tex]
