Question

very urgent friends please help me i have to submit it by tomorrow itself
Topic: Speed , Time And Distance

Question 1: express in meter/second: 108 km/hr

Question 2: express in km/hr:  15.5 m/s

Question 3: A Train Passes Through A Tunnel Whose Length Is 525 m , In 31 1/4 Seconds Moving At A Speed Of 72 km/hr . Find The Length Of The Train.

Question 4: Running At 4/3 Of His Usual Speed, A Person Improves His Timings By 10 Minutes.
Find His Usual Time To Cover The Distance.

Question 5: A Scooterist Covered A Distance In 10 Hours. He Covered The First Half Of The Distance At Speed Of 21 km/hr And The Other Half At A Speed Of 24 km/hr. Find The Distance He Travelled.

Answer 1

$108\ km/h = 108* \frac{5}{18}\ m/s =6*5\ m/s =30\ m/s\\ \\15.5\ m/s = 15.5* \frac{18}{5}\ km/h=3.1*18\ km/h=55.8\ km/h$

-----------------------------------------------------------------------------------
speed = 72 km/h = 72×(5/18) m/s = 20 m/s
time = 31 1/4 second = 31.25 seconds
distance covered = speed × time = 20 × 31.25 = 625m
length of tunnel = 525m
length of train = distance covered - tunnel length = 625 - 525 = 100m

---------------------------------------------------------------------------------
let the usual time taken to cover the distance = t minutes
If the speed increase by n times, the the time taken decreases by n times as they are inversely proportional.
so if speed becomes 4/3 times of usual, time taken will be 3/4 times of usual i.e. 3t/4. So
t - 3t/4 = 10
⇒ t/4 = 10
⇒ t = 4*10 = 40 minutes

------------------------------------------------------------------------------------
the speeds are 21 km/h and 24 km/h. Since he covered equal distance,

average speed = $\frac{2*21*24}{21+24}= 22.4$ km/h
time = 10 hours
distance = 22.4*10 = 224 km