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.
