Question

How much
required to bring the train to rest if it takes 5 sec to reduce its speed
from 90 kmph to 36 kmph: zs
(a) 5 s 25+. fb) 8.33 s
(c) 10.2 s (d) 12.43 s​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Time\:taken=8.34\:sec}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about initial and final velocity of a train and their time time interval is given.

• We have to find the time taken when final velocity become 0.

$\underline \bold{Given : } \\ \implies Initial \:velocity(u) = 90 \: km/h \\ \\ \implies Final \:velocity(u) = 36 \: km/h \\ \\ \implies Time \: taken(t) = 5 \: sec \\ \\ \underline \bold{To \: Find : } \\ \implies Time \: taken_{(at \: v = 0)} = ?$

• According to given question :

$\bold{u = 90 \times \frac{5}{18} = 25 \: m/s } \\ \\ \bold{v = 36 \times \frac{5}{18} = 10 \: m/s} \\ \\ \bold{Using \: first \: equation \: of \: motion : } \\ \implies v = u + at \\ \\ \implies10 = 25 + a \times 5 \\ \\ \implies a = \frac{ 10 - 25}{5} \\ \\ \implies a = \frac{\cancel{ - 15}}{\cancel5} \\ \\ \bold{\implies a = - 3 \: m/ {s}^{2} } \\ \\ \bold{For \: v = 0 : } \\ \implies v = u + at \\ \\ \implies 0 = 25 + (- 3) \times t \\ \\ \implies - 25 = - 3t \\ \\ \implies t = \frac{ - 25}{ - 3} \\ \\ \bold{\implies t = 8.34 \: sec}$

Answer 2

Answer:

8.33 sec :

Option b. is correct .

Explanation:

Given :

Variation in velocity :

90 km / hr and 36 km / hr .

In term of m / sec

Diving by 5 / 18

25 m / sec and 10 m / sec

Train is reducing its speed so we will take

u = 25 m / sec and v = 10 m / sec

Case 1.

Using first equation of motion :

a = 10 - 25 / 5

a = - 3

Case 2.

Now v = 0 and t' = ?

t' = - 25 / 3

t' = 8.33 sec

Hence we get time.