# Question baf29

Oct 29, 2017

#### Explanation:

Lets take initial distance to be d and time to be t.
So initially the speed (v₁) is:
${v}_{1} = \frac{d}{t}$

Then distance is halved and time is doubled. So the distance is now $\frac{d}{2}$ and time is $2 t$. What is our expression for speed (v₂)?
v_2 = (d//2)/(2t) = d/(4t) = ¼d/t#

Compare v₁ and v₂ :
${v}_{2} = {v}_{1} / 4$

So the answer is option 5.