Question

From a station 'X' a train starts from rest and attains a speed of 54 km/h in 10 s. Then by applying brakes negative acceleration of 2.5 m/s² is produced and it stops at station 'Y' in 6 s. Find the distance between 'X' and 'Y'.​

Answer 1

Answer:

Explanation:

First part of the motion,

u=0v=54 km/h=15 m/st=10 s

The acceleration,

v=u+at⇒a=vt=1510=1.5 m/s2

The distance travelled,

s1=ut+12at2⇒s1=12×1.5×102=75 m

In the second part,

t=8 min

the distance travelled,

s2=vt=15×8×60=7200 m

In the last part,

u=15 m/sv=0t=6 s

The deceleration,

v=u−at⇒a=ut/=15/6=2.5 m/s2

Distance travelled

v2=u2−2as⇒s3=u22a=1522×2.5=45 m

Distance between x and y=s1+s2+s3=75+7200+45=7320 m=7.32 km

Answer 2

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\pink{Solution:-}$

⏩ Here,

*u = 0,

*v= 54 km/h = 15 m/s,

*t = 10 s

We know that,

v= u + at

=> a= v/t

=>a = 15/10

=>a= 1.5 m/s²

→The distance travelled in first part,

s1= ut + 1/2 at²

= 0 + 1/2 × 1.5 × 10²

=75 m.

→In the last part,

u= 15 m/s, v= 0, t= 6s, a= 2.5 m/s²

Now,

v² = u² - 2as2

or,

s2 = u²/2a

= (15 m/s²)/ 2 × 2.5 m/s²

= 45 m

Therefore,

→ The Distance between 'X' and 'Y' = s1 + s2

= (75 +45) m

= 120 m. [The required solution..]

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