As a seasoned tutor registered on UrbanPro, I can confidently guide you through this problem. UrbanPro indeed provides excellent online coaching tuition services for students seeking academic assistance. Now, let's tackle your question.

When the jet airplane ejects its products of combustion, we have to consider the relative velocities to find the speed of the latter with respect to an observer on the ground. Here, the speed of the jet airplane is 500 km/h, and the speed of the ejected combustion products relative to the jet plane is 1500 km/h.

To find the speed of the combustion products relative to the ground observer, we need to apply the principle of relative velocity. We add the velocity of the jet airplane to the velocity of the ejected combustion products to find their combined velocity relative to the ground.

So, the speed of the combustion products relative to the ground observer would be: Speedground=Speedjet+Speedcombustion relative to jetSpeedground=Speedjet+Speedcombustion relative to jet

Speedground=500 km/h+1500 km/h=2000 km/hSpeedground=500 km/h+1500 km/h=2000 km/h

Therefore, the speed of the combustion products with respect to an observer on the ground would be 2000 km/h.

Feel free to reach out if you have any further questions or need clarification on any concept. That's what UrbanPro tutors are here for!

As an experienced tutor registered on UrbanPro, I can provide you with a step-by-step solution to this problem.

Let's break down the problem:

1. Initial Conditions: Both trains are moving in the same direction with Train A ahead of Train B. They both have a length of 400 meters each.

2. Speed and Acceleration: Both trains have a uniform speed of 71 km/h. Train B accelerates by 1 m/s² to overtake Train A.

3. Time Taken: After 50 seconds, the guard of Train B just brushes past the driver of Train A.

Now, let's calculate the distance between the two trains before Train B starts accelerating.

Since both trains are moving at the same speed, the relative speed at which Train B approaches Train A is the difference in their speeds, which is 71 km/h−71 km/h=0 km/h71km/h−71km/h=0km/h.

So, Train B's relative speed to Train A before acceleration is 0 km/h.

Given that the time taken for the guard of Train B to brush past the driver of Train A is 50 seconds, we can use the formula:

Distance=Speed×TimeDistance=Speed×Time

Since Train B's relative speed to Train A is 0 km/h, the distance covered by Train B during this time is 0 meters.

Therefore, the original distance between Train A and Train B before Train B started accelerating is simply the length of Train B, which is 400 meters.

So, the original distance between Train A and Train B was 400 meters.

As an experienced tutor registered on UrbanPro, I'd first like to commend your interest in understanding this physics problem. UrbanPro indeed offers a platform where students can find the best online coaching and tuition services to excel in their academic endeavors.

Now, let's tackle your physics question. We have three cars involved: A, B, and C. Car A is traveling at 36 km/h, while cars B and C are approaching A from opposite directions at 54 km/h each. At a certain point, when the distances AB and AC are both 1 km, B decides to overtake A before C does.

To avoid an accident, we need to ensure that B overtakes A safely without colliding with C. This means that B needs to complete the overtaking maneuver before C reaches the same point.

To calculate the minimum acceleration required for B to overtake A safely, we can use the equations of motion.

Let's denote:

• tt as the time taken for B to overtake A,
• aBaB as the acceleration of car B.

The distance covered by B during the overtaking maneuver is the sum of the initial distance between B and A (1 km) and the distance traveled by B while overtaking A.

The distance traveled by B during the overtaking maneuver can be calculated using the equation of motion: sB=uBt+12aBt2sB=uBt+21aBt2 where:

• sBsB is the distance traveled by B,
• uBuB is the initial velocity of B (54 km/h),
• aBaB is the acceleration of B, and
• tt is the time taken.

Given that the distance traveled by B must be equal to the initial distance between A and B (1 km), we can set up the equation as follows: 1+54t+12aBt2=11+54t+21aBt2=1

We need to solve this equation for aBaB, the acceleration of car B, to find the minimum acceleration required to avoid an accident.

• where g=9.8g=9.8 m/s² and h=90h=90 m.
• At t=2t=2 seconds, the ball has fallen for 2 seconds from rest, so its speed is v2=0.9×v1v2=0.9×v1.
• Repeat this process for t=3,4,5,...t=3,4,5,... until t=12t=12 seconds.

Once we have calculated the speed at each time interval, we can plot it on a graph with time on the x-axis and speed on the y-axis. This will give us the speed-time graph of the ball's motion.

If you're interested, I can provide the step-by-step calculations and then help you plot the graph. Let me know if you'd like to proceed!

Certainly! Let's break down the problem.

Given:

• Distance to market = 2.5 km
• Speed to market = 5 km/h
• Speed returning home = 7.5 km/h

(a) Magnitude of average velocity: Average velocity is the total displacement divided by the total time taken.

Total displacement = 0 km (since he returns home) Total time taken = Time to reach market + Time to return home

Time to reach market = Distance / Speed = 2.5 km / 5 km/h = 0.5 hours Time to return home = Distance / Speed = 2.5 km / 7.5 km/h = 1/3 hours

Total time taken = 0.5 hours + 1/3 hours = 5/6 hours

Average velocity = Total displacement / Total time taken = 0 km / (5/6) hours = 0 km/h

(b) Average speed: Average speed is the total distance traveled divided by the total time taken.

(i) 0 to 30 min (0.5 hours): Total distance traveled = 2.5 km (to market) + 2.5 km (returning home) = 5 km Average speed = Total distance / Total time = 5 km / 0.5 hours = 10 km/h

(ii) 0 to 50 min (5/6 hours): Total distance traveled remains the same (5 km) Average speed = Total distance / Total time = 5 km / (5/6) hours = 6 km/h

(iii) 0 to 40 min (2/3 hours): Total distance traveled = 2.5 km (to market) + 2.5 km (returning home) = 5 km Average speed = Total distance / Total time = 5 km / (2/3) hours = 7.5 km/h

UrbanPro is an excellent platform for online coaching and tuition, offering personalized guidance to help students understand and excel in their studies. If you need further clarification or assistance, feel free to ask!