Ch1 Ex 1d Q3 Two villages are 900 metres apart. 

The Problem:

Two villages are 900 metres apart. A car leaves the first village travelling at 15 m/s and accelerates at 1/2 m /s^2 for 30 seconds. How fast is it then travelling, and what distance has it covered in this time? The driver now sees the next village ahead, and decelerates so as to enter it at 15 m/s What constant deceleration is needed to achieve this? How much time does the driver save. by accelerating and decelerating, rather than covering the whole distance at 15 m/s?

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Douglas Quadling Mechanics1 Exercise1D Q3

Two villages are 900 metres apart. A car leaves the first village travelling at 15 ms and accelerates atm s2 for 30 seconds….

Second Method:

Car Traveling Between Two Villages

Car Traveling Between Two Villages

Given Data:

• Distance between villages: 900 meters
• Initial speed: 15 m/s
• Acceleration: 0.5 m/s² for 30 seconds
• Final speed before entering second village: 15 m/s

Step 1: Find Speed After 30 Seconds

Using the equation:

v = u + at

v = 15 + (0.5 × 30)

v = 15 + 15 = 30 m/s

Speed after 30 seconds is 30 m/s.

Step 2: Find Distance Covered in 30 Seconds

Using the equation:

s = ut + (1/2) a t²

s = (15 × 30) + (1/2 × 0.5 × 30²)

s = 450 + 225 = 675 meters

Distance covered in 30 seconds is 675 meters.

Step 3: Find Required Deceleration

Remaining distance to the second village:

900 – 675 = 225 meters

Using the equation:

v² = u² + 2as

15² = 30² + 2a(225)

225 = 900 + 450a

450a = -675

a = -1.5 m/s²

Required deceleration is 1.5 m/s².

Step 4: Find Time for Deceleration

Using the equation:

v = u + at

15 = 30 + (-1.5)t

t = 10 seconds

Time taken to decelerate is 10 seconds.

Step 5: Find Time Saved

(1) Time if the car moved at constant 15 m/s:

t = 900 / 15 = 60 seconds

(2) Time with acceleration and deceleration:

t_total = 30 + 10 = 40 seconds

(3) Time saved:

60 – 40 = 20 seconds

Final Answer

The driver saves 20 seconds by accelerating and decelerating instead of moving at a constant speed.

Third Method:

Two Villages Problem

Two villages are 900 meters apart. A car leaves the first village traveling at 15 m/s and accelerates at 0.5 m/s² for 30 seconds. How fast is it then traveling, and what distance has it covered in this time? The driver now sees the next village ahead and decelerates so as to enter it at 15 m/s. What constant deceleration is needed to achieve this? How much time does the driver save by accelerating and decelerating, rather than covering the whole distance at 15 m/s?

Solution

The problem is solved in three steps:

1. Speed and Distance During Acceleration:
   • Initial speed: 15 m/s.
   • Acceleration: 0.5 m/s².
   • Time accelerating: 30 seconds.
   • Final speed after accelerating:
     Final speed = Initial speed + (Acceleration × Time)
     Final speed = 15 + (0.5 × 30) = 15 + 15 = 30 m/s.
   • Distance covered during acceleration:
     Distance = (Initial speed × Time) + (0.5 × Acceleration × Time²)
     Distance = (15 × 30) + (0.5 × 0.5 × 30²) = 450 + 225 = 675 meters.
2. Deceleration Required to Enter the Second Village at 15 m/s:
   • Remaining distance to the second village:
     Total distance = 900 meters.
     Distance already covered = 675 meters.
     Remaining distance = 900 – 675 = 225 meters.
   • Required deceleration:
     The car slows down from 30 m/s to 15 m/s over 225 meters.
     Using the equation: Final speed² = Initial speed² + 2 × Deceleration × Distance
     Rearranged: Deceleration = (Final speed² – Initial speed²) / (2 × Distance)
     Deceleration = (15² – 30²) / (2 × 225) = (225 – 900) / 450 = -675 / 450 = -1.5 m/s².
     The negative sign indicates deceleration. So, the required deceleration is 1.5 m/s².
3. Time Saved by Accelerating and Decelerating:
   • Time taken with acceleration and deceleration:
     Time accelerating = 30 seconds.
     Time decelerating:
     Using the equation: Final speed = Initial speed + (Deceleration × Time)
     Rearranged: Time = (Final speed – Initial speed) / Deceleration
     Time = (15 – 30) / (-1.5) = (-15) / (-1.5) = 10 seconds.
     Total time = 30 + 10 = 40 seconds.
   • Time taken at constant speed (15 m/s):
     Total distance = 900 meters.
     Time = Distance / Speed = 900 / 15 = 60 seconds.
   • Time saved:
     Time saved = Time at constant speed – Time with acceleration and deceleration
     Time saved = 60 – 40 = 20 seconds.

Final Answers:

1. After accelerating for 30 seconds, the car is traveling at 30 m/s and has covered 675 meters.
2. The required deceleration to enter the second village at 15 m/s is 1.5 m/s².
3. The driver saves 20 seconds by accelerating and decelerating instead of traveling at a constant speed of 15 m/s.