Respuesta :
Answer:
1.41 m/s^2
Explanation:
First of all, let's convert the two speeds from km/h to m/s:
[tex]u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s[/tex]
[tex]v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s[/tex]
Now we find the centripetal acceleration which is given by
[tex]a_c=\frac{v^2}{r}[/tex]
where
v = 12.8 m/s is the speed
r = 140 m is the radius of the curve
Substituting values, we find
[tex]a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2[/tex]
we also have a tangential acceleration, which is given by
[tex]a_t = \frac{v-u}{t}[/tex]
where
t = 17.0 s
Substituting values,
[tex]a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2[/tex]
The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:
[tex]a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2[/tex]
Answer:
a = 1.406 m/s²
Explanation:
We are told that, the speed of the train decreases from 94.0 km/h to 46.0 km/h
Let's convert both to m/s.
Thus,
v1 =94 km/h =(94 x 10)/36 =26.11 m/s
v2=46 km/h =(46 x 10)/36=12.78 m/s
The formula to calculate the tangential acceleration is given by;
a_t = -dv/dt
Where;
dv is change in velocity
dt is time difference
dv is calculated as; dv = v1 - v2
Thus, dv = 26.11 - 12.78 = 13.33 m/s
We are given that, t = 17 seconds
Thus;
a_t = -13.33/17 = -0.784 m/s²
The negative sign implies that the acceleration is inwards.
Now, let's calculate the radial acceleration;
a_r = v²/r
Where;
r is the radius of the path = 140m
v is the velocity at the instant given
a_r is radial acceleration.
Thus,
a_r = 12.78²/140 = 1.167 m/s²
Now, the tangential and radial components of acceleration are perpendicular to each other. Thus, we can use using Pythagoreas theorem to find the resultant acceleration;
Thus;
a² = (a_t)² + (a_r)²
Plugging in the relevant values, we have;
a² = (-0.784)² + (1.167)²
a² = 1.976545
a = √1.976545
a = 1.406 m/s²
