Chapter 26, Problem 1RE

To determine

To decide: Whether the given statement is true or false.

Answer to Problem 1RE

The given statement is $\underset{\_}{\text{True}}$.

Explanation of Solution

Consider the give statement.

Let the initial velocity of an object be $v_0$.

Since the horizontal velocity of an object is $8{\text{ m/s}}^2$, the velocity along the straight line will be $v=v_0+8t$, where t is time.

Note that, the displacement s in terms of time t is $s={\displaystyle \int vdt}$.

Substitute $v=v_0+8t$ in $s={\displaystyle \int vdt}$ and simplify for s.

$\begin{array}{c}s={\displaystyle \int \left(v_0+8t\right)dt}\\ =v_0{\displaystyle \int tdt}+8{\displaystyle \int tdt}\\ =v_{0}t+4t^2+C\end{array}$

The distance travelled in previous second is obtained by replacing $t\text{ by }t-1$ as follows.

$\begin{array}{c}s\left(t-1\right)=v_0\left(t-1\right)+4{\left(t-1\right)}^2+C\\ =v_{0}t-v_0+4t^2+4-8t+C\end{array}$

Hence, distance travelled in the second before a given time is obtained as follows.

$\begin{array}{c}s\left(t\right)-s\left(t-1\right)=v_{0}t+4t^2+C-\left(v_{0}t-v_0+4t^2+4-8t+C\right)\\ =v_{0}t+4t^2+C-v_{0}t+v_0-4t^2-4+8t-C\\ =v_0+8t-4+\left(v_{0}t-v_{0}t\right)+\left(4t^2-4t^2\right)+\left(C-C\right)\\ =v_0+8t-4\end{array}$

The distance travelled in after the second is obtained by replacing $t\text{ by }t+1$ as follows.

$\begin{array}{c}s\left(t+1\right)=v_0\left(t+1\right)+4{\left(t+1\right)}^2+C\\ =v_{0}t+v_0+4t^2+4+8t+C\end{array}$

Hence, distance travelled in the second after a given time is obtained as follows.

$\begin{array}{c}s\left(t+1\right)-s\left(t\right)=\left(v_{0}t+v_0+4t^2+4+8t+C\right)-\left(v_{0}t+4t^2+C\right)\\ =v_{0}t+v_0+4t^2+4+8t+C-v_{0}t-4t^2-C\\ =v_0+8t+4+\left(v_{0}t-v_{0}t\right)+\left(4t^2-4t^2\right)+\left(C-C\right)\\ =v_0+8t+4\end{array}$

Calculate difference of the above distances.

$\begin{array}{c}s\left(t+1\right)-s\left(t\right)-s\left(t-1\right)+s\left(t\right)=v_0+8t+4-v_0-8t+4\\ s\left(t+1\right)-s\left(t-1\right)=8\end{array}$

Thus, the given statement is $\underset{\_}{\text{True}}$.