Solve Time Interval Between Two Cars Leaving with 0.4 m/s^2

In summary, the two cars traveled with an acceleration of 0.4 m/s^2 and after two minutes, the distance between them was 1.9 km. Using the suvat equations, it can be determined that the first car traveled 24 km in the first two minutes, while the second car traveled 24 * 120 - 1900 = 4780 km. The time interval between the departures of the cars can then be calculated using the equation s = ut + at^2/2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time interval.
  • #1
sambarbarian
68
0
Two cars leave one after the other and travel with acceleration 0.4 m/s^2 . Two minutes after the departure of the first vehicle , the distance between the cars becomes 1.9 km . The time interval between the departure of the cars is ?



i tried to do this by :- for First car (A)... 0.4 = Final V - 0 /120

therefore Final V is 48 m/s .

average V = 48 - 0 / 2 = 24

and 24 = d / 120 ... therefore distance = 24 * 120 .

now 24 * 120 - 1900 is the distance traveled by second car ( B ) ..

how should i proceed , if my logic is correct till now ?
 
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  • #2
You want kinematics ... look up the suvat equations and work in terms of distance traveled.

Break it down:
In the first 2min, how far has the first car travelled?
(You can do this directly from the time and the acceleration without referring to final velocity... but you are fine.)

How far has the second car travelled? (you've got this.)

How long has the second car been traveling to get there?
(Here is where you want that suvat equation then - you know s, u, and a; you want t.)

The time interval between departures.
 
  • #3
u mean S = ut + at^2/ 2 ??ok i got it , thank you .
 
  • #4
well done :)
 
  • #5


I would approach this problem by using the formula for distance travelled (d) with constant acceleration (a) and time (t): d = 1/2 * a * t^2. We know that the acceleration for both cars is 0.4 m/s^2 and the time interval between the cars is t. We also know that the distance between the two cars after two minutes is 1.9 km or 1900 m. Using this information, we can set up two equations:

For car A: d = 1/2 * 0.4 * t^2
For car B: d = 1/2 * 0.4 * (t+2)^2

We can then solve for t by setting the two equations equal to each other and solving for t:

1/2 * 0.4 * t^2 = 1/2 * 0.4 * (t+2)^2
0.2 * t^2 = 0.2 * (t+2)^2
0.2 * t^2 = 0.2 * (t^2 + 4t + 4)
0.2 * t^2 = 0.2 * t^2 + 0.8t + 0.8
0 = 0.8t + 0.8
t = -1

However, this solution does not make sense in the context of the problem. We know that the time interval between the cars cannot be negative. This means that there is likely an error in the initial information or calculations. I would suggest double checking the values and equations used to ensure accuracy and make any necessary corrections.
 

Related to Solve Time Interval Between Two Cars Leaving with 0.4 m/s^2

1. What is the formula for calculating time interval between two cars leaving with 0.4 m/s^2?

The formula for calculating time interval between two cars leaving with 0.4 m/s^2 is t = (v2 - v1)/a, where t is the time interval, v2 is the final velocity, v1 is the initial velocity, and a is the acceleration.

2. Can the time interval be negative in this scenario?

No, the time interval cannot be negative in this scenario. The formula for calculating time interval accounts for the difference between the final and initial velocities, so the result will always be a positive value.

3. How do I determine the direction of acceleration in this problem?

The direction of acceleration can be determined by looking at the initial and final velocities of the cars. If the final velocity is greater than the initial velocity, the acceleration is in the same direction as the motion of the cars. If the final velocity is less than the initial velocity, the acceleration is in the opposite direction.

4. What does the value of 0.4 m/s^2 represent in this problem?

The value of 0.4 m/s^2 represents the acceleration of the cars. This means that the speed of the cars is increasing by 0.4 meters per second every second.

5. Can this formula be used for any scenario involving two objects with different initial and final velocities?

Yes, this formula can be used for any scenario involving two objects with different initial and final velocities, as long as the acceleration remains constant.

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