Physics, asked by annonymous27, 9 months ago

A train starts from rest and accelerates uniformly at 100 m/min2 for 10 minutes. Find the velocity acquired by the train. It then maintains a constant velocity for 20 minutes. The brakes are then applied and the train is uniformly retarded. It comes to rest in 5 minutes. Draw a velocity-time graph and use it to find: i) the retardation in the last 5 minutes ii) total distance travelled, and iii) the average velocity of the train

Answers

Answered by krishnarajpoot8293
0

Answer:

Let ‘u’ be initial velocity.

Let ‘v’ be final velocity.

The final velocity of the train is:

\bold{v = u + at}v=u+at

\bold{v = 0 + (100 \times 10)}v=0+(100×10)

v = 1000 m/s

So, this is the maximum velocity of the train.

The distance covered is given by the formula:

\bold{S = \frac{1}{2}at^2}S=

2

1

at

2

Distance covered during uniform acceleration:

\bold{S_{1} = \frac{1}{2}at^{2}_{1}}S

1

=

2

1

at

1

2

\bold{S_{1} = \frac{1}{2} \times 100 \times 10^{2}}S

1

=

2

1

×100×10

2

\bold{S_{1} = 5000 \ m = 5 \ km}S

1

=5000 m=5 km

Distance covered during constant velocity:

\bold{S_{2} = v \times t_{2}}S

2

=v×t

2

\bold{S_{2} = 1000 \times 20}S

2

=1000×20

\bold{S_{2} = 20000 \ m = 20 \ km}S

2

=20000 m=20 km

Distance covered during retardation:

Retardation = \bold{200 \ m/min^{2}}200 m/min

2

\bold{S_{3} = \frac{1}{2}at^{2}_{3}}S

3

=

2

1

at

3

2

\bold{S_{3} = \frac{1}{2} \times 200 \times 5^{2}}S

3

=

2

1

×200×5

2

\bold{S_{3} = 2500 \ m = 2.5 \ km}S

3

=2500 m=2.5 km

Now, the total distance is:

\bold{S = S_{1} + S_{2} + S_{3}}S=S

1

+S

2

+S

3

\bold{S = 5 + 20 + 2.5}S=5+20+2.5

\bold{\therefore S = 27.5 \ km}∴S=27.5 km

Answered by dhruvsharma81
1

Explanation:

Total distance covered(s) = Distance during acceleration(s

1

) + distance during uniform motion(s

2

) + distance during retardation(s

3

)

For acceleration,

v=u+a∗t

v

max

=2×10=20 m/s

s=ut+(1/2)at

2

s

1

=(1/2)×2×10

2

=100 m

For uniform motion,

s=vt

s

2

=20×200

=4000 m

During retardation,

v=u+at

0=20+50t

=−0.4 m/s

2

s

3

=ut+(1/2)at

2

=20×50+(1/2)×(−0.4)×50

2

=500 m

Total distance travelled, s=s1+s2+s3=100+4000+500=4600 m

Total time taken, t=t

1

+t

2

+t

3

=260 s

Average velocity, v

avg

=

Total Time

Total Distance

=4600/260=17.69 m/s

Similar questions