Question

A space shuttle moving at $2\text{m/s}$ at $t=0$ begins to accelerate at a constant rate of $2{\text{m/s}}^2$. If $S_n$ is the distance covered by it in the ${\text{n}}^{\text{th}}$ second, then:

• A. $S_1+S_4>S_2+S_3$
• B. $S_1+S_4=S_2+S_3$
• C. $S_1+S_4<S_2+S_3$

Answer 1

The correct option is B $S_1+S_4=S_2+S_3$

We know,
$S_n=u\left(1\right)+\frac{1}{2}a(2n-1)$
Also, it is given that,
$u=2\text{m/s}$
$a=2{\text{m/s}}^2$
Replacing values in the equation we get
$\Rightarrow S_n=2+(2n-1)$
$\Rightarrow S_n=2n+1$
$\therefore S_1=3\text{m}$
$S_2=5\text{m}$
$S_3=7\text{m}$
$S_4=9\text{m}$

$\Rightarrow S_1+S_4=S_2+S_3=12\text{m}$