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

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

                             where m is the slope

Comparing equation (1) and (2)

a = m    

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

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

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

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.

= -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:

where;

= the distance traveled up

= 0

= speed of the train

= deceleration

t = time

Substituting our data; we have:

= 123.297 m

where are the vectors ?