Question

A body covers a distance of 4 m in $3^{rd}$ second and 12 m in $5^{th}$ second. If the motion is uniformly accelerated, how far will it travel in the next 3 seconds?

• A. 10 m
• B. 30 m
• C. 40 m
• D. 60 m

Answer 1

Answer: (D)

60 m

We know that distance Travelled by uniformly accelerated body in $n^{th}s$ is given by

$S_n=u+\frac{1}{2}a(2n-1),$ where $a$ is the $acceleration$ of the body and $u$ is the initial velocity

it is given that $S_n=4s$ at $n=3$

$⟹4=u+\frac{5a}{2}$ .............. $eq\left(1\right)$

$S_n=12$ at $n=5s$

$⟹12=u+\frac{9a}{2}$ ............... $eq\left(2\right)$

solving $eq\left(1\right)$ and $eq\left(2\right),$ we get

$a=4m/s^2$ and $u=-6m/s$

Now we have to find velocity at $t=5s$

we know that $v=u+at$

$⟹v=-6+(4\times 5)=14m/s$

now distance travelled in next $3s$ can be calculated by

$S=ut+\frac{1}{2}a{t}^2$ $Note:$ now $u=14m/s$

$⟹S=14\times 3+\frac{1}{2}4\left(3^2\right)=60m$