At t is equal to zero and x is equal to zero and initial stationery bl...
**Problem Statement**
A blue car starts from rest and accelerates at a constant rate of 0.2 m/s² in the positive direction of the x-axis. At t = 2 seconds, a red car traveling in an adjacent lane and in the same direction passes x = 0 with a speed of 8.0 m/s and constant acceleration of 3.0 m/s². Find the time at which the red car passes the blue car.
**Solution**
To solve the problem, we need to determine the position of both cars as a function of time and then find the time at which their positions are equal. We can use the following kinematic equations to determine the position of each car:
- x = x₀ + v₀t + ½at²
where x is the position of the car at time t, x₀ is the initial position, v₀ is the initial velocity, a is the acceleration, and t is the time.
**Calculations for Blue Car**
- Given: x₀ = 0, v₀ = 0, a = 0.2 m/s²
- x = 0 + 0t + ½(0.2)t²
- x = 0.1t²
**Calculations for Red Car**
- Given: x₀ = 0, v₀ = 8.0 m/s, a = 3.0 m/s²
- x = 0 + 8.0t + ½(3.0)t²
- x = 8.0t + 1.5t²
**Finding the Time at which Red Car Passes Blue Car**
To find the time at which the red car passes the blue car, we need to set their positions equal to each other and solve for t.
- 0.1t² = 8.0t + 1.5t²
- 0.4t² - 8.0t = 0
- t(0.4t - 8.0) = 0
This equation has two solutions: t = 0 and t = 20 seconds. The first solution corresponds to the time when both cars are at x = 0, which is not the time when the red car passes the blue car. Therefore, the time at which the red car passes the blue car is:
- t = 20 seconds
**Conclusion**
In conclusion, the red car will pass the blue car after 20 seconds. At this time, the position of the blue car is x = 0.1(20)² = 40 meters, and the position of the red car is x = 8.0(20) + 1.5(20)² = 320 meters. Therefore, the red car passes the blue car when they are both at x = 40 meters.
At t is equal to zero and x is equal to zero and initial stationery bl...
Is it t= 2.05s (appx) ?
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