Answer:
Acceleration is -1.2 m/s² and distance covered is 135 m.
Explanation:
A train going at a speed of 18m / s brakes and stops in 15s calculates its acceleration and the distance traveled when braking
Given that,
The initial speed of the train, u = 18 m/s
Final speed, v = 0
Time, t = 15 s
We need to find acceleration and distance traveled when braking. Let a is acceleration and distance traveled.
Acceleration,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18}{15}\\\\a=-1.2\ m/s^2[/tex]
Using third equation of motion,
[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-18^2}{2\times (-1.2)}\\\\d=135\ m[/tex]
Hence, acceleration is -1.2 m/s² and distance covered is 135 m.
