Finding the Distance Traveled in the 10th Second of a Truck's Deceleration

This problem involves understanding motion under constant acceleration, specifically deceleration, and calculating the distance covered during a specific time interval, the 10th second.

Step 1: Convert Initial Velocity

The initial velocity of the truck is given in kilometers per hour (km/hr). We need to convert this to meters per second (m/s) for consistency with standard units used in physics calculations.

Initial velocity, $u = 54 \text{ km/hr}$.

To convert km/hr to m/s, we use the conversion factor $\frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18}$.

$u = 54 \times \frac{5}{18} \text{ m/s}$

$u = 3 \times 5 \text{ m/s}$

$u = 15 \text{ m/s}$.

Step 2: Calculate Deceleration (Acceleration)

The truck is brought to rest, meaning its final velocity is 0 m/s. This happens over a time period of 10 seconds. We can use the first equation of motion, $v = u + at$, to find the acceleration ($a$). Since the truck is slowing down, the acceleration will be negative (deceleration).

• Initial velocity, $u = 15 \text{ m/s}$
• Final velocity, $v = 0 \text{ m/s}$
• Time taken, $t = 10 \text{ s}$

Using the equation:

$v = u + at$

$0 = 15 + a \times 10$

$-15 = 10a$

$a = \frac{-15}{10} \text{ m/s}^2$

$a = -1.5 \text{ m/s}^2$.

The acceleration is $-1.5 \text{ m/s}^2$, indicating a deceleration of $1.5 \text{ m/s}^2$. This constant deceleration is key to finding the distance traveled.

Step 3: Calculate Distance in the 10th Second

To find the distance traveled specifically in the 10th second, we can use two methods:

Method 1: Difference of Distances

The distance traveled in the 10th second is the difference between the total distance traveled in 10 seconds and the total distance traveled in the first 9 seconds.

We use the second equation of motion, $s = ut + \frac{1}{2}at^2$, to find the distance traveled in a given time $t$. Here, $u = 15 \text{ m/s}$ and $a = -1.5 \text{ m/s}^2$.

Distance traveled in 10 seconds ($S_{10}$):

$S_{10} = ut + \frac{1}{2}at^2$ (with $t=10$)

$S_{10} = (15)(10) + \frac{1}{2}(-1.5)(10)^2$

$S_{10} = 150 + \frac{1}{2}(-1.5)(100)$

$S_{10} = 150 - 0.75 \times 100$

$S_{10} = 150 - 75$

$S_{10} = 75 \text{ m}$.

Distance traveled in 9 seconds ($S_9$):

$S_9 = ut + \frac{1}{2}at^2$ (with $t=9$)

$S_9 = (15)(9) + \frac{1}{2}(-1.5)(9)^2$

$S_9 = 135 + \frac{1}{2}(-1.5)(81)$

$S_9 = 135 - 0.75 \times 81$

$S_9 = 135 - 60.75$

$S_9 = 74.25 \text{ m}$.

Distance traveled in the 10th second ($S_{10th}$):

$S_{10th} = S_{10} - S_9$

$S_{10th} = 75 \text{ m} - 74.25 \text{ m}$

$S_{10th} = 0.75 \text{ m}$.

Method 2: Formula for Distance in $n^{th}$ Second

The distance traveled in the $n^{th}$ second can be directly calculated using the formula:

$S_{n^{th}} = u + \frac{a}{2}(2n - 1)$

Here, $u = 15 \text{ m/s}$, $a = -1.5 \text{ m/s}^2$, and we want the distance in the 10th second, so $n = 10$.

$S_{10^{th}} = 15 + \frac{-1.5}{2}(2 \times 10 - 1)$

$S_{10^{th}} = 15 + (-0.75)(20 - 1)$

$S_{10^{th}} = 15 + (-0.75)(19)$

$S_{10^{th}} = 15 - (0.75 \times 19)$

$S_{10^{th}} = 15 - 14.25$

$S_{10^{th}} = 0.75 \text{ m}$.

Both methods yield the same result for the distance traveled in the 10th second of the truck's motion.

Conclusion

The distance traveled by the truck in the 10th second is 0.75 m.

Revision Table: Key Concepts in Kinematics

Additional Information on Motion Problems

Understanding motion problems often relies on correctly identifying the given parameters and selecting the appropriate kinematic equation. For motion with constant acceleration, there are three primary equations relating initial velocity ($u$), final velocity ($v$), displacement ($s$), acceleration ($a$), and time ($t$).

• $v = u + at$ (Velocity-Time relation)
• $s = ut + \frac{1}{2}at^2$ (Position-Time relation)
• $v^2 = u^2 + 2as$ (Velocity-Position relation)

When a body is slowing down, the acceleration is negative, often called deceleration or retardation. Converting units correctly (like km/hr to m/s) is a crucial first step in solving physics problems. The distance in the $n^{th}$ second concept helps find the displacement specifically during that single second, not the total displacement up to that time.