Question

An object starting from rest travels 20 m in first 2 seconds and 160 m in next 4 seconds. What will be the velocity after 7 seconds from the start?

Answer 1

Step 1: Given data:

As the object starts from rest;
Initial velocity($u$) = 0 m/s
Distance($s$) travelled in first $2s$ = $20m$
Distance travelled($S$) in next $4s$ = $160m$

Step 2: Formula used:

The first and second equation of motion is given as;

$v=u+at$

$s=ut+\frac{1}{2}a{t}^2$

Where, u= Initial velocity of the body

v= final velocity of the body

a= Acceleration in the body

t= Time taken by the body

s= Distance covered by the body

Step 3: Finding acceleration for the first 2 sec in the object:

Let $\text{a}$ be the acceleration of the object,

Using the second equation of motion for the first 2 sec of the motion of the object;

$20m=0\times 2s+\frac{1}{2}\times a\times {\left(2s\right)}^2$

$20=\frac{1}{2}\times a\times 4$

$\Rightarrow \frac{1}{2}\times a\times 4=20$

$\Rightarrow 2a=20$

$\Rightarrow a=10m{s}^{-2}$

Step 4: Finding the final velocity of the object after the first 2 sec:

Let, v be the velocity of the body after 2sec; then using the first equation of motion,

$v=0+10m{s}^{-2}\times 2s$

$v=20m{s}^{-1}$

Step 5: Finding acceleration in the object for the next 4 sec:

For the next 4 sec, the final velocity of the body in 2 sec will be the initial velocity of the object.

That is for the next 4 sec, initial velocity= $v=10m{s}^{-1}$

Let a' be the acceleration in the object for the next 4sec.
So, using the second equation of motion;

$160m=20m{s}^{-1}\times 4sec+\frac{1}{2}\times a^{\prime }\times {\left(4sec\right)}^2$

$160=80+8a^{\prime }$

$\Rightarrow 80+8a^{\prime }=160$

$\Rightarrow 8a^{\prime }=80$

$\Rightarrow a^{\prime }=10m{s}^{-2}$

It is seen that both a and a' are equal thus, the object is moving with uniform acceleration.

Step 6: Calculation of the velocity after 7sec after start:

At start, u= 0

Therefore let the velocity of the object is V then, using the first equation of motion, we get;

$V=0+10m{s}^{-2}\times 7sec$

$V=70m{s}^{-1}$

Hence,

The velocity after 7 sec from the start is 70 m/s.