Unit 4: Time, Speed and Moving Objects - Subjective Questions

Definitions:

• Distance: The total length of the path travelled by an object.
• Time: The duration taken by an object to complete a journey.
• Speed: The distance travelled by an object per unit of time.

The fundamental relationship is:

Rearranging it gives:

and

These formulas are directly applicable when the object moves at a uniform speed. If the speed changes during the journey, average speed or segment-wise calculations must be used.

Since km m and hour seconds:

Therefore:

• To convert km/h into m/s, multiply by .
• To convert m/s into km/h, multiply by .

First conversion:

Second conversion:

Thus, km/h is m/s, and m/s is km/h.

The relationship is:

For a constant distance, time is inversely proportional to speed:

Let the original speed be and the original time be . After a increase, the new speed is:

Therefore, the new time is:

The time becomes of the original time. Hence, the time taken decreases by .

Average speed is the ratio of the total distance travelled to the total time taken:

It is generally not equal to the arithmetic mean of the speeds because the object may spend different amounts of time at different speeds.

Let each equal distance be km. Then:

The total distance is . Therefore:

For equal distances at speeds and , the average speed is the harmonic mean:

Thus, the required average speed is km/h, not km/h.

Let the one-way distance be km.

The onward and return times are:

Since the total time is hours:

Taking the LCM:

The total distance is km. Therefore:

• One-way distance: km
• Average speed: km/h

Let the distance be km. The original speed is km/h, and the increased speed is km/h.

The difference in travel times is hours:

Using the LCM :

The original travel time is:

At the increased speed, the time would be:

Therefore, the distance is km and the original travel time is hours.

Let the speeds of the cars be km/h and km/h.

The difference in travel times is minutes, or hour:

Therefore:

• Slower car's speed: km/h
• Faster car's speed: km/h

A check gives times of hours and hours, whose difference is minutes.

When A completes m, B covers m. Therefore:

When B completes m, C covers m. Therefore:

Combining the ratios:

Thus, when A covers m, the distance covered by C is:

Hence, A defeats C by:

Therefore, A defeats C by m.

When A runs m, B runs m. Hence:

When A runs m, C runs m. Hence:

Therefore:

When B runs m, C runs:

Thus, B can give C a start of:

Therefore, B can give C a start of m.

Relative speed is the speed at which the distance between two moving objects changes.

Objects moving in the same direction:

If their speeds are and , where , then:

The faster object gains on the slower object at the difference of their speeds.

Objects moving in opposite directions:

If their speeds are and , then:

The distance between them changes at the sum of their speeds.

Relative speed converts a two-object motion problem into an equivalent one-object problem, making it useful in questions involving trains, races, meetings, overtaking, and circular tracks.

Since the trains move in opposite directions, their relative speed is the sum of their speeds.

Convert the speeds into m/s:

Thus:

To cross completely, they must cover the sum of their lengths:

Therefore:

Hence, the trains cross each other completely in approximately seconds.

The train must cover its own length relative to the walking man.

The relative speed is:

Convert the man's speed into m/s:

Since the train and the man move in the same direction:

Therefore:

Converting into km/h:

Hence, the speed of the train is km/h.

Since the cyclists move towards each other, their relative speed is:

The time required to meet is:

The distance travelled by the cyclist starting from A is:

The distance travelled by the cyclist starting from B is:

The distances add to km, confirming the calculation.

Therefore, they meet after hours, at a point km from A or km from B.

Because the runners move in the same direction, their relative speed is:

The faster runner catches the slower runner after gaining one complete lap, or m.

Distance covered by the faster runner:

Number of laps completed by the faster runner:

Distance covered by the slower runner:

Number of laps completed by the slower runner:

Thus, the faster runner catches the slower runner after seconds. They have completed laps and laps respectively.

Let the speed of a boat in still water be and the speed of the stream be .

• Downstream motion: The boat moves in the direction of the stream, so the stream assists it.

• Upstream motion: The boat moves against the stream, so the stream reduces its effective speed.

Adding the two equations:

Therefore:

Subtracting the upstream speed from the downstream speed:

Therefore:

Thus, the speed in still water is the average of the downstream and upstream speeds, while the stream speed is half their difference.

The downstream speed is:

The upstream speed is:

The speed of the boat in still water is:

The speed of the stream is:

Therefore:

• Speed of the boat in still water: km/h
• Speed of the stream: km/h

Let the speed of the boat in still water be km/h.

Then:

According to the question:

Combining the fractions:

Since speed must be positive:

The upstream and downstream speeds are km/h and km/h. Their travel times are hours and hours, differing by hours. Hence, the boat's speed in still water is km/h.

Let the distances travelled by train and bus be km and km respectively.

From the total distance:

The time taken by train is hours, and the time taken by bus is hours. Therefore:

Multiplying the second equation by :

Multiplying the first equation by :

Subtracting gives:

Substituting into :

Therefore, the person travels km by train and km by bus.

In the first hours, the first car gains a lead of:

Once the second car starts, both cars move in the same direction. Their relative speed is:

The time required for the second car to close the km gap is:

The distance travelled by the second car during this time is:

The first car travels for a total of hours:

Thus, the second car overtakes the first hours after the second car starts, or hours after the first car starts, at a point km from A.

Let the distance be km and the scheduled travel time be hours.

At km/h, the man is minutes, or hour, late:

At km/h, he is minutes, or hour, early:

Subtracting the second equation from the first:

Using the first equation:

Therefore, the distance is km and the scheduled travel time is hours.