Uniform Speed, Variable Speed, and Conversions

Imagine strolling steadily on a treadmill versus playing tag, where you speed up and slow down. One is constant; the other keeps changing!

• Distance (m): How far an object moves.

• Time (s): How long the movement takes.

• Speed (m/s): How fast an object moves.

• Uniform Speed: Covers equal distances in equal times.

• Variable Speed: Covers different distances in equal times.

Average speed of an object = `"Total distance covered" / "Total time taken"`

To convert between m/s and km/h:

• multiply by 18/5 (m/s to km/h)
  
  
• multiply by 5/18 (km/h to m/s)

Examples:

• 90 km/h = $25$ m/s

• 15 m/s = $54$ km/h

• Passing a pole or any other stationary object, etc.: Distance = Length of train

• Passing a platform: Distance = Length of train + Length of platform

Problem: A man runs 200 metres in 25 seconds. 
Find:

(i) his speed. 
(ii) the distance run by him in 5 seconds
(iii) the time taken by him to cover `2/5` km.

Solution:

(i) Speed = `"Distance" / "Time "` = `"200 m" / "25 sec"` = 8 m s⁻¹ 

(ii) Distance run in 5 sec = Speed × Time
                                        = 8 m s⁻¹ × 5 sec
                                        = 40 m    

(iii) Time taken to cover `2/5` km = `"Distance" / "Time "`

                                                      =  `400 / (8 m s^-1)`

                                                      = 50 seconds 

Problem: A train covers the first 120 km in 2 hours, the next 160 km in 3 hours, and the last 140 km again in 2 hours. Find the average speed of the train.

Solution:

Average speed of an object = `"Total distance covered" / "Total time taken". `

Since the total distance covered = 120 km + 160 km + 140 km 

                                              = 420 km
Total time taken = 2 hr + 3 hr + 2 hr

                         = 7 hr.

Therefore, Average speed = `"420km"/"7hr"`

                                          = 60 km h⁻¹

Convert:

(i)  90 km h⁻¹ into m s⁻¹.

90 km h⁻¹ = 90 × `5/8` ms⁻¹ = 25 ms⁻¹

(ii) 15 ms⁻¹ into km h⁻¹

15 m s⁻¹ = 15 × `18/5` km h⁻¹ = 54 km h⁻¹

(iii) 75 cm s⁻¹ into km h⁻¹

75 cm s⁻¹ = 0.75 m s⁻¹

= 0.75  × `18/5` km h⁻¹ = 2.7 km h⁻¹ 

Problem: A 160 m long train is travelling at a speed of 72 km/h. Find the time taken by the train to pass:

(i) a telegraph post 
(ii) a 200 m long platform. 

Solution:

(i) Distance to be covered = length of the train = 160 m
And speed = 72 km h⁻¹ = 72 × m s⁻¹ = 20 m s⁻¹

∴ Time taken = `"Distance"/"Speed"` = `160 /20` sec = 8 sec 

(ii) Distance to be covered = length of the train + length of the platform 

                                           = 160 m + 200 m = 360 m 

Time taken = `"Distance"/"Speed"` = `360 /20` sec = 18 sec 

• Uniform speed: same distance in same time.

• Variable speed: distance changes each time.

• Speed = `"Distance" / "Time "`

• Average speed of an object = `"Total distance covered" / "Total time taken"`