To make this problem solvable and you can get the help you need, I'll complete and arrange some data.
Answer:
Acceleration: [tex]0.0417\ m/s^2[/tex], Distance=7,500 m
Explanation:
Uniform Acceleration Motion
It's a type of motion in which the velocity of an object changes uniformly over time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
[tex]v_f=v_o+at\qquad\qquad [1][/tex]
The distance traveled by the object is given by:
[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2][/tex]
Using the equation [1] we can solve for a:
[tex]\displaystyle a=\frac{v_f-v_o}{t}\qquad\qquad [3][/tex]
The problem will be rewritten as follows:
A train starting from rest reaches a velocity of 90 km/h in 10 minutes. Assuming that the acceleration is uniform, find the acceleration and the distance traveled by the train for attending the velocity.
Let's take the relevant data:
vo=0
vf=90 Km/h*1000/3600 = 25 m/s
t = 10 minutes = 10*60 = 600 seconds
Now compute the acceleration by using [3]:
[tex]\displaystyle a=\frac{25-0}{600}=0.0417[/tex]
[tex]a=0.0417\ m/s^2[/tex]
Finally, compute the distance:
[tex]\displaystyle x=0*600+\frac{0.0417\cdot 600^2}{2}[/tex]
[tex]x=7,500\ m[/tex]
Note: We used the value of the acceleration with more precision than shown.
Acceleration: [tex]\mathbf{0.0417\ m/s^2}[/tex], Distance=7,500 m
