Respuesta :
What is the average acceleration of a subway train that speeds up from 9.6 m/s to 12 m/s in 0.8 s on a straight section of track? Show your work below.
Average acceleration = change in velocity/ change in time
=12-9.6/.8=2.4/.8=3m/s²
Average acceleration = change in velocity/ change in time
=12-9.6/.8=2.4/.8=3m/s²
Answer:
Average acceleration, [tex]a=3\ m/s^2[/tex]
Step-by-step explanation:
Given that,
Initial speed of the train, u = 9.6 m/s
Final speed of the train, v = 12 m/s
Time taken, t = 0.8 s
Let a is the acceleration of the train. The mathematical formula for acceleration is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{12-9.6}{0.8}[/tex]
[tex]a=3\ m/s^2[/tex]
So, the average acceleration of the subway train is [tex]3\ m/s^2[/tex]. Hence, this is the required solution.