Question

Obtain a relation for the distance travelled by an object moving with uniform acceleration in the Interval between 4^th and 5^th seconds.

Answer 1

Step 1: Given that:

The object is moving with uniform acceleration.

Step 2: Formula used:

The distance covered by a body in a given time $t$ under uniform acceleration is given by;

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

Where, $s$ = Distance travelled by the body

$u$ = Initial velocity of the body

$a$ = Uniform acceleration in the body

Step 3: Calculation of the distance travelled by the object in the Interval between 4th and 5th seconds:

If $u$ be the initial velocity of the body and uniform acceleration of the body is taken as $a$ then,
Distance travelled by the object in $5s$ will be given as;

$s_5=u\times 5sec+\frac{1}{2}a\times {5sec}^2$

$s_5=5u+\frac{25}{2}a$..............(1)

The distance covered by the body in $4sec$ will be given as;

$s_4=u\times 4sec+\frac{1}{2}a\times {\left(4sec\right)}^2$

$s_4=4u+\frac{16}{2}a$

$s_4=4u+8a$............(2)

Now,

The distance travelled by the body in the interval between 4th and 5th second is given by the difference between the distance travelled in $5sec$ and $4sec$.

Thus,

Required distance= $s_5-s_4$

Now,

Putting the values of $s_{5}and{s}_4$ we get,

$=5u+\frac{25}{2}a-(4u+8a)$

$=5u+\frac{25}{2}a-4u-8a$

$=u+\left(\frac{25-16}{2}\right)a$

$=(u+\frac{9}{2}a)m$

Thus,

The distance covered by the body in the Interval between 4th and 5th seconds will be $(u+\frac{9}{2}a)m$ .