Answer:
123.30 m
Explanation:
Given
Speed, u = 22 m/s
acceleration, a = 1.40 m/s²
time, t = 7.30 s
From equation of motion,
v = u + at
where,
v is the final velocity
u is the initial velocity
a is the acceleration
t is time
V = at + U
using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane
[tex]V_{x} = mt + V_{o}[/tex] where m is the slope
Comparing equation (1) and (2)
[tex]V = V_{x}[/tex]
a = m
[tex]U = V_{o}[/tex]
Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.
a = - 1.40 m/s²
Sunstituting a = - 1.40 m/s² and u = 22 m/s
[tex]V_{x} = -1.40t + 22[/tex]
[tex]V_{x} = -1.40(7.30) + 22[/tex]
[tex]V_{x} = -10.22 + 22[/tex]
[tex]V_{x} = 11. 78 m/s[/tex]
The speed of the train at 7.30 s is 11.78 m/s.
The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.
Area of triangle + Area of rectangle
[tex][\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)][/tex]
= 37.303 + 85.994
= 123. 297 m
≈ 123. 30 m
Answer:
123.297 m
Explanation:
A train, traveling at a constant speed of 22.0 m/s,
v = 22.0 m/s
the train slows down with a constant acceleration of magnitude 1.40 m/s².
[tex]a_s[/tex] = -1.4 m/s²
How far has the train traveled up the incline after 7.30 s
t =7.30 s
We can calculate the distance traveled up the incline after 7.30 s by using the formula:
[tex]x_f =x_i+v_xt+\frac{1}{2}a_st^2[/tex]
where;
[tex]x_f[/tex] = the distance traveled up
[tex]x_i[/tex] = 0
[tex]v_x[/tex] = speed of the train
[tex]a_s[/tex] = deceleration
t = time
Substituting our data; we have:
[tex]x_f = 0+(22m/s)(7.30s)+\frac{1}{2}(-1.4m/s^2)(7.30s)[/tex]
[tex]x_f =16.06 -37.303[/tex]
[tex]x_f[/tex] = 123.297 m
