Respuesta :
Answer:
total distance = 1868.478 m
Explanation:
given data
accelerate = 1.68 m/s²
time = 14.2 s
constant time = 68 s
speed = 3.70 m/s²
to find out
total distance
solution
we know train start at rest so final velocity will be after 14 .2 s is
velocity final = acceleration × time ..............1
final velocity = 1.68 × 14.2
final velocity = 23.856 m/s²
and for stop train we need time that is
final velocity = u + at
23.856 = 0 + 3.70(t)
t = 6.44 s
and
distance = ut + 1/2 × at² ...........2
here u is initial velocity and t is time for 14.2 sec
distance 1 = 0 + 1/2 × 1.68 (14.2)²
distance 1 = 169.37 m
and
distance for 68 sec
distance 2= final velocity × time
distance 2= 23.856 × 68
distance 2 = 1622.208 m
and
distance for 6.44 sec
distance 3 = ut + 1/2 × at²
distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²
distance 3 = 76.90 m
so
total distance = distance 1 + distance 2 + distance 3
total distance = 169.37 + 1622.208 + 76.90
total distance = 1868.478 m
Answer:
1868.5 m
Explanation:
For AB :
u = 0 m/s
a = + 1.68 m/s^2
t = 14.2 s
Let the distance is s1 and the velocity at B is v.
Use first equation of motion
v = u + at
v = 0 + 1.68 x 14.2 = 23.856 m/s
Use third equation of motion
[tex]v^{2}=u^{2}+2as_{1}[/tex]
[tex]23.856^{2}=0^{2}+2\times1.68\times s_{1}[/tex]
s1 = 169.38 m
For BC:
Let the distance is s2.
s2 = v x t
s2 = 23.856 x 68 = 1622.21 m
For CD:
u = 23.856 m/s
a = - 3.7 m/s^2
v = 0
Let the distance is s3.
Use third equation of motion
[tex]v^{2}=u^{2}+2as_{3}[/tex]
[tex]0^{2}=23.856^{2}-2\times 3.7 \times s_{3}[/tex]
s3 = 76.91 m
The total distance traveled is
s = s1 + s2 + s3
s = 169.38 + 1622.21 + 76.91 = 1868.5 m
Thus, the total distance traveled is 1868.5 m.
