Question

PLZZ solve this problem...​

Answer 1

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✒ v = 20 m/s

✒ a = - 0.4 m/s²

✒ s = 4600 m

✒ Avv = 17.69 m/s

$\bold{\Huge\green{\boxed{{{Given}}}}}$

✒ Accelerates 2 m/s in 10 second

✒ Constant speed for 200 second

✒ Come in rest in 50 second

$\bold{\Huge\blue{\boxed{{{To \: Find}}}}}$

1. The maximum velocity reached.

2. Retardation in last 50 seconds.

3. Total distance travelled.

4. The average velocity of train.

$\bold{\huge\pink{\boxed{{{Explanation}}}}}$

⭐First of all we have 3 cases/values in question. (we need them later)

* t1= 10 , u1 = 0 , v1 = 20 , a = 2

* t2= 200 , v = v1 = 20 , a = 0

* t3= 50 , u2 = v = 20 , v2=0 , a = - 0.4

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1) We know that, a = v-u/t

So, Substituting the values

2 = v - 0 / 10

v = 20 m/s (Above values) is the

maximum velocity reached.

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2) We know that, a = v-u/t

So,bSubstituting the values

a = 0 - 20 / 50

a = - 0.4 m/s² is the acceleration in last 50 seconds.

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3) Total Distance travelled = Distance in

case 1 + case 2 + case 3

✏ case 1. (using 3rd eq. Of motion)

=> 2as = v² - u²

= 2(2)s = (20)² - (0)²

= 4s = 400

= s = 400÷4

=> s = 100 m

✏ case 2. (using 2nd eq. Of motion)

=> s = ut + ½ at²

= s = (20)(200) + ½ (0)(200)²

= s = 4000 + 0

=> s = 4000 m

✏ case 3. (using 3rd eq. Of motion)

=> 2as = v² - u²

= 2(-0.4)s = (0)² - (20)²

= (-0.8)s = - 400

= s = 4000÷8

=> s = 500 m

Now, (100 + 4000 + 500) m

=> 4600 m is the total distance.

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4) Average velocity = Toatal Displacement/Total time (s/t>

*Total Displacement = 4600

*Total Time = 260 second

So, Substituting the values

=> 4600÷260

=> 17.69 m/s is the average velocity of train.

$\bold{\Huge{\boxed{{{</u></strong><strong><u>Note</u></strong><strong><u>}}}}}$

⭐Avv = Average Velocity

⭐ u = Initial velocity

⭐ v = Final velocity

⭐ a = Retardation (here use)

⭐ t = Time

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Answer 2

Answer:

Total distance covered(a) = Distance during acceleration(s) + distance during uniform motion(sz) + distance during retardation(sz)

For acceleration,

v = u +at

Vmax = 2 x 10 = 20 m/s

s=ut + (1/2)at?

S1 = (1/2) x 2 x 10²

= 100 m

For uniform motion,

S = vt

S2 = 20 x 200

= 4000 m

During retardation, v = u + at

0= 20 + 50t

= -0.4 m/s2

S3 = ut + (1/2)at?

= 20 x 50 + (1/2) x (-0.4) x 502

= 500 m

Total distance travelled, s = sl + s2 + s3 =

100 + 4000 + 500 4600 m

Total time taken, t=tı + t2 + t3 = 260 s

Average velocity, Vavg =

Total Distance

Total Time

= 4600/260 = 17.69 m/s

Explanation:

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