A body starts from rest, what is the ratio of the distance travelled by the during the $4^{th}$ and $3^{rd}$ second ?

Question

A car starts from rest and accelerates uniformly at 3 m/s\textsuperscript{2. What will be its velocity after 5 seconds?}

Answer 1

To solve this problem, we need to calculate the distances traveled during the 3rd and 4th seconds of the motion of a body starting from rest under uniform acceleration. We can use the equations of motion for uniformly accelerated linear motion.

Since the body starts from rest, initial velocity $u = 0$ and let acceleration be $a$. The distance traveled in the $n^{th}$ second is given by:

$$ s_n = u + \frac{1}{2} a (2n - 1) $$

Since the initial velocity $u = 0$, this equation simplifies to:

$$ s_n = \frac{1}{2} a (2n - 1) $$

Now, using this formula, we calculate the distances traveled during the 3rd and 4th seconds:

Distance during the 3rd second, s_3:

$$ s_3 = \frac{1}{2} a (2 \cdot 3 - 1) = \frac{1}{2} a \cdot 5 = \frac{5}{2} a $$

Distance during the 4th second, s_4:

$$ s_4 = \frac{1}{2} a (2 \cdot 4 - 1) = \frac{1}{2} a \cdot 7 = \frac{7}{2} a $$

Now, we find the ratio of the distance traveled in the 4th second to the distance traveled in the 3rd second:

$$ \text{Ratio} = \frac{s_4}{s_3} = \frac{\frac{7}{2} a}{\frac{5}{2} a} = \frac{7}{5} $$

Thus, the ratio of the distance traveled during the 4th second to that during the 3rd second is $\frac{7}{5}$, which matches the given correct answer.

Answer 2