Until a train is a safe distance from the station, it must travel at [tex][tex]$5 \, \text{m/s}$[/tex][/tex]. Once the train is on open track, it can speed up to [tex][tex]$45 \, \text{m/s}$[/tex][/tex]. If it takes the train 8 seconds to reach [tex][tex]$45 \, \text{m/s}$[/tex][/tex], what is the acceleration of the train? (Round your answer to the nearest whole number.)

A. [tex]4 \, \text{m/s}^2[/tex]
B. [tex]5 \, \text{m/s}^2[/tex]
C. [tex]6 \, \text{m/s}^2[/tex]
D. [tex]7 \, \text{m/s}^2[/tex]



Answer :

To find the acceleration of the train, we can use one of the basic kinematic equations that relates acceleration to the initial speed, final speed, and time. The formula for acceleration [tex]\(a\)[/tex] is given by:

[tex]\[ a = \frac{v_f - v_i}{t} \][/tex]

where:
- [tex]\( v_f \)[/tex] is the final speed,
- [tex]\( v_i \)[/tex] is the initial speed,
- [tex]\( t \)[/tex] is the time over which the change in speed occurs.

In this problem:
- The initial speed ([tex]\( v_i \)[/tex]) of the train is [tex]\( 5 \)[/tex] meters per second,
- The final speed ([tex]\( v_f \)[/tex]) is [tex]\( 45 \)[/tex] meters per second,
- The time ([tex]\( t \)[/tex]) taken to reach this final speed is [tex]\( 8 \)[/tex] seconds.

Substituting these values into the formula, we get:

[tex]\[ a = \frac{45 \, \text{m/s} - 5 \, \text{m/s}}{8 \, \text{s}} \][/tex]

[tex]\[ a = \frac{40 \, \text{m/s}}{8 \, \text{s}} \][/tex]

[tex]\[ a = 5 \, \text{m/s}^2 \][/tex]

Thus, the calculated acceleration is [tex]\( 5 \, \text{m/s}^2 \)[/tex].

Since the problem asks us to round the answer to the nearest whole number, we see that the calculated acceleration is already a whole number.

Therefore, the rounded acceleration of the train is [tex]\( 5 \, \text{m/s}^2 \)[/tex], which corresponds to the answer:

[tex]\[ \boxed{5 \, \text{m/s}^2} \][/tex]