Question::

a motorcyclist moving with uniform retardatin takes 10s and 20s to travel sucessive quarter kilometer. how much further he will travel before coming to rest?

I AM CONFUSED WITH QUARTERLY PLS CLEAR THIS CONCEPT.

To tackle this problem, let’s break it down step by step. The motorcyclist is moving with uniform retardation, which means he is slowing down at a constant rate. The key here is to understand how distance, time, and acceleration (or in this case, deceleration) relate to each other. The term "quarter kilometer" simply means 250 meters, as a kilometer is 1000 meters. So, the motorcyclist travels 250 meters in the first 10 seconds and another 250 meters in the next 20 seconds. Let’s analyze this situation further.

Understanding the Motion

We know the motorcyclist covers two distances of 250 meters each, but the time taken for each segment is different. The first segment takes 10 seconds, and the second takes 20 seconds. This indicates that the motorcyclist is decelerating, as he takes longer to cover the same distance in the second segment.

Calculating the Deceleration

To find out how much further the motorcyclist will travel before coming to a stop, we first need to determine his initial speed and the rate of deceleration. We can use the equations of motion for uniformly accelerated (or decelerated) motion:

• First Segment: Distance (s1) = 250 m, Time (t1) = 10 s
• Second Segment: Distance (s2) = 250 m, Time (t2) = 20 s

We can find the average speed during each segment:

• Average speed for the first segment: v1 = s1 / t1 = 250 m / 10 s = 25 m/s
• Average speed for the second segment: v2 = s2 / t2 = 250 m / 20 s = 12.5 m/s

Finding the Deceleration

Now, we can find the deceleration (a) using the formula:

a = (v2 - v1) / t

Here, we can consider the change in speed over the time taken to travel from the first segment to the second:

a = (12.5 m/s - 25 m/s) / 10 s = -1.25 m/s²

Calculating the Distance Before Coming to Rest

Now that we have the deceleration, we can find out how much further the motorcyclist will travel before coming to a stop. We can use the formula:

v² = u² + 2as

Where:

• v = final velocity (0 m/s, since he comes to rest)
• u = initial velocity (which is the speed at the end of the second segment, 12.5 m/s)
• a = acceleration (which is -1.25 m/s²)
• s = distance traveled before coming to rest

Plugging in the values:

0 = (12.5)² + 2(-1.25)s

This simplifies to:

0 = 156.25 - 2.5s

Rearranging gives:

2.5s = 156.25

Thus:

s = 156.25 / 2.5 = 62.5 m

Final Thoughts

In conclusion, the motorcyclist will travel an additional 62.5 meters before coming to a complete stop. Understanding the relationship between distance, time, and speed is crucial in solving problems involving motion, especially when dealing with uniform acceleration or deceleration. If you have any further questions or need clarification on any part of this, feel free to ask!