Question

Два мотоциклиста выезжают одновременно из двух пунктов навстречу друг другу. Один из них спускается равноускоренно с горы, имея начальную скорость 36 км/ч и ускорение 2 м/с2. Другой равнозамедленно поднимается в гору с начальной скоростью 72км/ч и с тем же по модулю ускорением. первоначальное расстояние между мотоциклистами равно 300 м. Через сколько времени они встретятся?

Answer 1

Problem Analysis

We have two motorcyclists starting from two different points and moving towards each other. One is descending a hill with an initial speed of 36 km/h and an acceleration of 2 m/s^2, while the other is ascending a hill with an initial speed of 72 km/h and the same acceleration. We need to find out how long it will take for them to meet.

Solution

To solve this problem, we can use the equations of motion. Let's assume that the time taken for the motorcyclists to meet is 't'.

For the motorcyclist descending the hill: - Initial velocity (u) = 36 km/h - Acceleration (a) = 2 m/s^2 - Distance covered (s) = ?

For the motorcyclist ascending the hill: - Initial velocity (u) = 72 km/h - Acceleration (a) = -2 m/s^2 (negative because it is decelerating) - Distance covered (s) = ?

We know that the distance covered can be calculated using the equation: s = ut + (1/2)at^2

For the motorcyclist descending the hill: s1 = (36 km/h) * (t hours) + (1/2) * (2 m/s^2) * (t^2 hours^2)

For the motorcyclist ascending the hill: s2 = (72 km/h) * (t hours) + (1/2) * (-2 m/s^2) * (t^2 hours^2)

Since the motorcyclists are moving towards each other, the sum of their distances covered should be equal to the initial distance between them, which is 300 m. s1 + s2 = 300 m

Let's solve this equation to find the value of 't'.

Calculation

Let's convert the initial velocities from km/h to m/s: - Initial velocity of the descending motorcyclist = 36 km/h = (36 * 1000) / 3600 m/s = 10 m/s - Initial velocity of the ascending motorcyclist = 72 km/h = (72 * 1000) / 3600 m/s = 20 m/s

Now, let's substitute the values into the equation: (10t + (1/2) * 2 * t^2) + (20t + (1/2) * (-2) * t^2) = 300

Simplifying the equation: 10t + t^2 + 20t - t^2 = 300 30t = 300 t = 300 / 30 t = 10 seconds

Answer

The two motorcyclists will meet after 10 seconds.

Verification

Let's verify the answer using the given information.

For the motorcyclist descending the hill: - Initial velocity (u) = 36 km/h = 10 m/s - Acceleration (a) = 2 m/s^2 - Time (t) = 10 seconds

Using the equation of motion: s1 = ut + (1/2)at^2 s1 = (10 m/s) * (10 s) + (1/2) * (2 m/s^2) * (10 s)^2 s1 = 100 m + 100 m s1 = 200 m

For the motorcyclist ascending the hill: - Initial velocity (u) = 72 km/h = 20 m/s - Acceleration (a) = -2 m/s^2 - Time (t) = 10 seconds

Using the equation of motion: s2 = ut + (1/2)at^2 s2 = (20 m/s) * (10 s) + (1/2) * (-2 m/s^2) * (10 s)^2 s2 = 200 m - 100 m s2 = 100 m

The sum of the distances covered by the motorcyclists is 200 m + 100 m = 300 m, which matches the initial distance between them. Therefore, the answer is verified.

Note: The search results provided by You.com did not contain any relevant information for this specific problem. The solution provided above is based on the principles of kinematics and the equations of motion.