Question

15. A train starts from rest and accelerates uniformly at
a rate of 2 m s-2 for 10 s. It then maintains a
constant speed for 200 s. The brakes are then
applied and the train is uniformly retarded and
comes to rest in 50 s. Find : (i) the maximum
velocity reached, (ii) the retardation in the last
50 s, (iii) the total distance travelled, and (iv) the
average velocity of the train.
-1 G 0.4​

Answer 1

ANSWER:

1. The maximum velocity reached = 20 m/s
2. The retardation in the last = 0.4 m/s²
3. The total distance travelled = 4600 m
4. The average velocity of the train = 17.69 m/s

GIVEN:

• The train travels at 2m/s² for 10 s

• Starts at rest

• After 10s, It travels at a constant speed for 200 s

• Retards uniformly for 50 s and comes to rest after 210 s.

TO FIND:

1. The maximum velocity reached
2. The retardation in the last
3. The total distance travelled
4. The average velocity of the train.

EXPLANATION:

1) The maximum velocity reached:

The maximum velocity should be reached after 10 seconds, because after that it travels at constant speed and again retards.

u = 0

a = 2 m/s²

t = 10 s

$\large{ \bf{ \boxed{ \sf{v = u + at}}}}$

v = 0 + 2(10)

v = 20 m/s

The maximum velocity will be 20 m/s.

2) The retardation in the last:

The initial velocity is what we found above because after 10 s the train travels at a constant speed, then starts retarding and at last it comes to rest(v = 0).

u = 20 m/s

v = 0

t = 50s

$\large{ \bf{ \boxed{ \sf{v = u + at}}}}$

0 = 20 + a(50)

50a = - 20

a = - 2/5

a = - 0.4 m/s²

It retards uniformly at a rate of 0.4 m/s².

3) The total distance travelled:

Lets split the distance as three parts A,B,C

First part:

$\large{ \bf{ \boxed{ \sf{s = ut + \dfrac{1}{2}at^2}}}}$

A = 0(10) + 1/2(2)(10²)

A = 1/2(2 × 100)

A = 100 m

Second part:

Here velocity is constant and hence accleration is zero

$\large{ \bf{ \boxed{ \sf{Distance = Speed \times Time }}}}$

B = 20 × 200

B = 4000 m

Third part:

$\large{ \bf{ \boxed{ \sf{s = ut + \frac{1}{2}a {t}^{2} }}}}$

C = 20(50) + 1/2(- 0.4)(50)(50)

C = 1000 - (2500)/5

C = 1000 - 500

C = 500 m

${\bf{\boxed {\sf{Total \: \: distance \: = A + B + C}}}}$

Total distance = 100 + 4000 + 500

Total distance = 4600 m.

4) The average velocity of the train:

${\bf{\boxed {\sf{Avg \: \: velocity = \dfrac{Total \: \: distance}{Total \: \: time}}}}}\\ \\ \\ \sf{Avg \: \: velocity = \dfrac{4600}{{260}}}\\ \\ \\ \sf{Avg \: \: velocity = \dfrac{100 + 4000 + 500}{{10 + 200 + 50}}}$

Avg velocity = 17.69 m/s.

Answer 2

Answer:

17.69 m/s

Explanation:

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