Question

A body $A$ starts from rest with an acceleration $a_1$. After$2seconds$, another body $B$ starts from rest with an acceleration$a_2$. If they travel equal distances in the$5thsecond$, after the start of $A$, then the ratio$a_1:a_2$ is equal to

• A. $5:9$
• B. $5:7$
• C. $9:5$
• D. $9:7$

Answer 1

The correct option is A

$5:9$

Step 1. Given data:

Initial velocity is $u=0$

Acceleration of body $A$ is $a_1$

Acceleration of body $B$ is $a_2$

Distance travelled by both body are equal, that's $S_A=S_B$

Step 2. Calculating distance covered in $n^{th}second$ using Newton's equation of motion:

From $2^{nd}$equation of motion, distance travelled in t second, $s=ut+\frac{1}{2}a{t}^2$

Distance travelled in $nsecond$, $S_n=un+\frac{1}{2}a{n}^2$ ____ $\left(1\right)$

Distance travelled in $\left(n-1\right)second$, $S_{\left(n-1\right)}=u\left(n-1\right)+\frac{1}{2}a{\left(n-1\right)}^2$ ____$\left(2\right)$

We can calculate distance travelled in $n^{th}$second as follows

$$\begin{gathered} S_n^{th}=S_n-S_{\left(n-1\right)} \\ {S}_n^{th}=un+\frac{1}{2}a{n}^2-\left[u\left(n-1\right)+\frac{1}{2}a{\left(n-1\right)}^2\right] \end{gathered}$$

On simplifying the above equation, we get

Distance covered in the nth second after starting from rest, ,$S_{nth}=u+\frac{1}{2}a(2n-1)$

$S_{nth}=\frac{a}{2}\left(2n-1\right)\left(Givenu=0\right)$

Step 3. Calculating distance covered by both body in terms of their acceleration:

For $A$, $n=5anda=a_1$

So, $S_A=\frac{a}{2}\left(2n-1\right)$

$=\left(\frac{a_1}{2}\right)(2x5–1)$

$=\left(\frac{9}{2}\right)a1$

For $B$, $n=3anda=a_2$ (Body B starts $2seconds$ after the body A starts)

So, $S_B=\frac{a}{2}\left(2n-1\right)$

$$\begin{gathered} =\frac{a_2}{2}\left(2\times 3-1\right) \\ =\frac{5a_2}{2} \end{gathered}$$

Step 4. Equating distance covered by both body:

As given, $S_A=S_B$

$\Rightarrow \left(\frac{9}{2}\right)a_1=\left(\frac{5}{2}\right)a_2$

$\Rightarrow a_1:a_2=5:9$

Hence, Option A is correct.