Answer:
1796.48 m
Explanation:
Velocity after 14 seconds can be given by the equation,
Final Velocity=Initial Velocity + Acceleration*Time
v=0+1.6×14=22.4m/s^2
Distance covered during these 14 seconds,
s1=(a×t^2)/2=1.6×14×14/2=156.8
Distance during the next 70 seconds,
s2=22.4×70=1568 (using the formula s=u×t+(1/2)at^2)
Time taken to stop, use the equation v=u+at
0=22.4-3.5×t
t=6.4 seconds
Distance traveled during these 6.4seconds,
s3=22.4×6.4-(1/2)3.5×6.4×6.4=143.36-71.68=71.68
Total distance = s1+s2+s3=1796.48m
The total distance covered by the subway train is calculated as 1798.18m
Data;
To solve this question effectively, we need to draw the velocity-time graph.
The distance can be calculated by adding the sum of the individual area of the figures.
Area of first triangle.
[tex]A = \frac{1}{2}b*h\\[/tex]
But the height here would be the velocity of this train.
[tex]velocity = acceleration * time\\velocity = 1.6*14=22.4m/s[/tex]
The height of the triangle is 22.4m
Let's substitute the values and solve for area
[tex]A_1 = \frac{1}{2} * 14 * 22.4 = 157.5m[/tex]
The next distance we would find is the area of the rectangle
[tex]A = L*W\\l= 22.4\\w = 70\\A = 22.4*70=1568m[/tex]
And the last distance we need to calculate is the last triangle
Area of triangle is
[tex]A = \frac{1}{2}*b*h\\[/tex]
But we don't know the value of the base (time) it took for the deceleration.
Let's use the velocity-acceleration-time relationship
[tex]acceleration = \frac{velocity}{time}\\ time = \frac{velocity}{acceleration}\\ time = \frac{22.4}{3.5}\\ time = 6.4s[/tex]
Substitute the values and solve for the Area
[tex]A= \frac{1}{2}b*h\\A = \frac{1}{2}* 6.4*22.4\\A = 72.68m[/tex]
The total distance covered by the train is 157.5m + 1568m + 72.68m
[tex]distance=157.5+1568+72.68\\distance=1798.18m[/tex]
The total distance covered by the subway train is calculated as 1798.18m
Learn more on velocity-time graph here;
https://brainly.com/question/14086970
