Question

A body covers 4m in the 3rd second and 12m in the 5th second . If the motion is uniformly accelerated . How far will it travel the next 3second

Answer 1

Task. A body covers 4m in the 3rd second and 12m in the 5th second. If the motion is uniformly accelerated, how far will it travel the next 3 second.

Solution. Let $v$ be the initial velocity of the body and $a$ be its acceleration. Then the distance covered by the body at time $t$ is given by the following formula:

$d(t)=vt+\frac{gt^{2}}{2}.$

We should find the distance covered by the body at next 3 seconds after 5 seconds, i.e. at time $t=5+3=8$ s.

By assumption

$d(3\ s)=4\ m$

and

$d(5\ s)=12\ m.$

Thus

$4=d(3)=v*3+\frac{g*3^{2}}{2}=3v+4.5g$

and

$12=d(5)=v*5+\frac{g*5^{2}}{2}=5v+12.5g$

So we obtain the following system of equations

\[ \left\{\begin{array}[]{rcl}3v+4.5g&=&4\\

5v+12.5g&=&12\end{array}\right.\Rightarrow\left\{\begin{array}[]{rcl}6v+9g&=&8\\

10v+25g&=&24\end{array}\right. \]

Multiplying both sides of the first equation by $(-5)$, both sides of the second equation by 3 and adding them we obtain

\[ \left\{\begin{array}[]{rcl}-30v-45g&=&-40\\

30v+75g&=&72\end{array}\right. \]

$-30v-45g+30v+75g=72-40$

$30g=32$

$g=\frac{32}{30}=\frac{16}{15}\approx 1.0667\ m/s^{2}.$

Therefore

$3v=4-4.5g$

and so

$v=\frac{4-4.5g}{3}=\frac{4-\frac{9}{2}\cdot\frac{16}{15}}{3}=\frac{4-\frac{3*8}{5}}{3}=\frac{4-4.8}{3}=-\frac{0.8}{3}=-\frac{8}{30}=-\frac{4}{15}\approx-0.26667\ m/s.$

Hence

$d(8)=-\frac{4}{15}*8+\frac{16}{15}*\frac{8^{2}}{2}=-\frac{32}{15}+\frac{16*32}{15}=\frac{16*32-32}{15}=\frac{15*32}{15}=32\ m.$

Answer. 32 $m$