High School

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------------------------------------------------ A train starts at the station at rest and begins accelerating at a rate of 5.9 m/s². After 38.1 seconds, how far has the train traveled?

Answer :

Answer:

4282.2m is the distance travelled by the train.

Explanation:

The illustration can be represented using velocity - time graph attached.

Let v represents its final velocity.

To calculate the value of the final velocity v, find the slope of the v-t graph, acceleration is the slope of a velocity -time slope.

[tex]slope = acceleration \: = \frac{change \: in \: velocity}{time} [/tex]

[tex]acceleration= \frac{v - 0}{t - 0} [/tex]

Given the acceleration = 5.9m/s² and time = 38.1 s ,

[tex]5.9= \frac{v - 0}{38.1 - 0} [/tex]

[tex]5.9= \frac{v }{38.1 }[/tex]

Multiply both sides by 38.1

v = 38.1 × 5.9

v = 224.79 m/s

Thus, the final velocity is 224.79 m/s.

How to find the distance travelled by the

training during the time.

Distance travelled by the train = Area of the triangle under the graph

Distance travelled by the train =

[tex] \frac{1}{2}bh[/tex]

where:

  • b = time = 38.1
  • h = final velocity = 224.79

Therefore,

Distance travelled by the train

[tex] = \frac{1}{2} \times 38.1 \times 224.79[/tex]

= 4282.2495

= 4282.2m to the nearest tenth

Answer:

Explanation:

To find out how far the train has traveled after 38.1 seconds, we can use the kinematic equation for displacement under constant acceleration:

=

0

+

1

2

2

d=v

0

t+

2

1

at

2

where:

d is the displacement,

0

v

0

is the initial velocity,

a is the acceleration,

t is the time.

Given:

Initial velocity,

0

=

0

m/s

v

0

=0m/s (the train starts at rest),

Acceleration,

=

5.9

m/s

2

a=5.9m/s

2

,

Time,

=

38.1

s

t=38.1s.

Since the initial velocity

0

=

0

v

0

=0, the equation simplifies to:

=

1

2

2

d=

2

1

at

2

Substituting the given values:

=

1

2

×

5.9

×

(

38.1

)

2

d=

2

1

×5.9×(38.1)

2

Calculating:

=

0.5

×

5.9

×

1451.61

d=0.5×5.9×1451.61

=

2.95

×

1451.61

d=2.95×1451.61

4282.25

m

d≈4282.25m

Conclusion

The train has traveled approximately 4,282.3 meters after 38.1 seconds.