Answer:
[tex] \boxed{\sf Acceleration \ of \ train = 1 \ m/s^{2}} [/tex]
Given:
Initial speed (u) = 0 km/hr = 0 m/s
Final speed (v) = 36 km/hr
Time taken (t) = 10 sec
To Find:
Acceleration (a) of the train
Explanation:
[tex]\sf 1 \ km/hr = \frac{5}{18} \ m/s \\ \therefore \\ \sf 36 \ km/hr = 36 \times \frac{5}{18} \ m/s \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2 \times \cancel{18} \times \frac{5}{ \cancel{18}} \ m/s \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2 \times 5 \ m/s \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 10 \ m/s[/tex]
So,
Final speed (v) = 10 m/s
[tex]\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ t: \\ \sf \implies 10 = 0 + a(10) \\ \sf \implies 10 = a(10) \\ \sf \implies a \times 10 = 10 \\ \\ \sf Dividing \ both \ sides \ by \ 10: \\ \sf \implies \frac{a \times 10}{\boxed{\sf 10}} = \frac{10}{\boxed{\sf 10}} \\ \\ \sf \frac{\cancel{10}}{\cancel{10}} = 1: \\ \sf \implies a = 1 \: m/s^2[/tex]
Acceleration of train = 1 m/s²
