Question

A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the train slows down with a constant acceleration of magnitude 1.40 m/s2. How far has the train traveled up the incline after 7.30 s

Answer 1

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.

$a_s$ = -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:

$x_f =x_i+v_xt+\frac{1}{2}a_st^2$

where;

$x_f$ = the distance traveled up

$x_i$ = 0

$v_x$ = speed of the train

$a_s$ = deceleration

t = time

Substituting our data; we have:

$x_f = 0+(22m/s)(7.30s)+\frac{1}{2}(-1.4m/s^2)(7.30s)$

$x_f =16.06 -37.303$

$x_f$ = 123.297 m

Answer 2

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

                       $V_{x} = mt + V_{o}$      where m is the slope

Comparing equation (1) and (2)

$V = V_{x}$

a = m    

$U = V_{o}$

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

                        $V_{x} = -1.40t + 22$

                            $V_{x} = -1.40(7.30) + 22$

                             $V_{x} = -10.22 + 22$

                             $V_{x} = 11. 78 m/s$

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

          $[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]$

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m