Question

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

Answer 1

Step 1: Given data

1. The acceleration of the body is constant (uniform acceleration) with time.
2. Time of travel of the body is from the interval 4th to 5th sec.

Step 2: Displacement

1. Displacement is a vector quantity. It is the minimum distance of the path of travel.
2. From the concept of kinematics, displacement is defined by the form, $S=ut+\frac{1}{2}a{t}^2$, where, u is the initial velocity of a body, a is the acceleration and t is the time of travel.

Step 3: Finding the distance

Using the formulae, $S=ut+\frac{1}{2}a{t}^2$,

the distance traveled by the body in 5 seconds is,

$$\begin{gathered} S_5=\left(u\times 5\right)+\frac{1}{2}a\times \left(5^2\right) \\ orS=5u+\frac{25}{2}a.......................\left(1\right) \end{gathered}$$

Similarly, the distance traveled by the body in 4 seconds is,

$$\begin{gathered} S_4=\left(u\times 4\right)+\frac{1}{2}a\times \left(4^2\right) \\ orS=4u+\frac{16}{2}a \\ orS=4u+8a.......................\left(2\right) \end{gathered}$$

Now, the distance traveled by the body in the interval 4^th and 5^th seconds is,

$$\begin{gathered} S=\left(S_5-S_4\right)=\left(5u+\frac{25}{2}a\right)-\left(4u+8a\right) \\ orS=\left(5u-4u\right)+\left(\frac{25}{2}a-8a\right) \\ orS=u+\frac{9}{2}a \end{gathered}$$

Therefore, the distance traveled by the body in the interval between 4^th and 5^th seconds is $S=u+\frac{9}{2}a$. This is the relation.