Two cars A & B are coming closer to one another with speed 36 km/h and 72 km/h
Separated by a distance 1500 m calculate :
1 . time interval after which two car meet .
2 . if acceleration of car A is 2m/s2 then calculate time interval after which car A & B
Will meet.
3.if acceleration of car A is 2m/s2 and of car B is 3 m/s2 then time interval
after which car A & B will meet.
To solve these problems, we can use the equation for calculating the time taken for two objects moving towards each other to meet:
\[ \text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} \]
1. Time interval after which two cars meet:
The relative speed of the cars is the sum of their individual speeds, as they are moving towards each other.
\[ \text{Relative Speed} = 36 \text{ km/h} + 72 \text{ km/h} = 108 \text{ km/h} \]
Converting the relative speed to meters per second (m/s):
\[ \text{Relative Speed} = 108 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 30 \text{ m/s} \]
Using the equation, \[ \text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} \]
\[ \text{Time} = \frac{1500 \text{ m}}{30 \text{ m/s}} = 50 \text{ seconds} \]
Therefore, the two cars will meet after 50 seconds.
2. Time interval after which car A and B will meet, given the acceleration of car A is 2 m/s²:
If the acceleration of car A is given, we can use the equation of motion to calculate the time it takes for car A to meet car B.
The equation is:
\[ \text{Distance} = \text{Initial Velocity} \times \text{Time} + \frac{1}{2} \times \text{Acceleration} \times \text{Time}^2 \]
In this case, we know the initial velocity of car A is 36 km/h, and the distance is 1500 m.
We can first convert velocity to m/s:
\[ \text{Initial Velocity} = 36 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 10 \text{ m/s} \]
Rearranging the equation to solve for time:
\[ \text{Time} = \frac{-\text{Initial Velocity} \pm \sqrt{\text{Initial Velocity}^2 + 2 \times \text{Acceleration} \times \text{Distance}}}{\text{Acceleration}} \]
Simplifying,
\[ \text{Time} = \frac{-10 \text{ m/s} \pm \sqrt{(10 \text{ m/s})^2 + 2 \text{ m/s}^2 \times 1500 \text{ m}}}{2 \text{ m/s}^2} \]
Evaluating this equation, we find two possible solutions for time. Taking the positive value,
\[ \text{Time} = \frac{-10 \text{ m/s} + \sqrt{(10 \text{ m/s})^2 + 2 \text{ m/s}^2 \times 1500 \text{ m}}}{2 \text{ m/s}^2} \approx 33.75 \text{ seconds} \]
Therefore, car A and B will meet after approximately 33.75 seconds.
3. Time interval after which car A and B will meet, given the accelerations of car A and car B:
Using the same equation of motion as before, we need to consider the effects of both accelerations.
The equation becomes:
\[ \text{Distance} = \text{Initial Velocity} \times \text{Time} + \frac{1}{2} \times \text{Acceleration}_\text{Total} \times \text{Time}^2 \]
Here, the acceleration is the sum of the accelerations of both cars:
\[ \text{Acceleration}_\text{Total} = 2 \text{ m/s}^2 + 3 \text{ m/s}^2 = 5 \text{ m/s}^2 \]
Using the equation, solving for time:
\[ \text{Time} = \frac{-\text{Initial Velocity} \pm \sqrt{\text{Initial Velocity}^2 + 2 \times \text{Acceleration}_\text{Total} \times \text{Distance}}}{\text{Acceleration}_\text{Total}} \]
As before, evaluating this equation, we find two possible solutions for time. Taking the positive value,
\[ \text{Time} = \frac{-10 \text{ m/s} + \sqrt{(10 \text{ m/s})^2 + 2 \text{ m/s}^2 \times 1500 \text{ m}}}{5 \text{ m/s}^2} \approx 7.75 \text{ seconds} \]
Therefore, car A and B will meet after approximately 7.75 seconds, given the different accelerations.