Question

A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round the bend. The radius of the curve is 140 m. Compute the acceleration at the moment the train speed reaches 46.0 km/h. Assume the train continues to slow down at this time at the same rate.

Answer 1

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

$u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s$

$v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s$

Now we find the centripetal acceleration which is given by

$a_c=\frac{v^2}{r}$

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

$a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2$

we also have a tangential acceleration, which is given by

$a_t = \frac{v-u}{t}$

where

t = 17.0 s

Substituting values,

$a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2$

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

$a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2$

Answer 2

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