A train applies it brakes slows down from a speed of 15.0 m/s to a stop in 30.0 s. The distance it travels before it stops is _____ m.
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Use equation of motion
v = u + at
Rearrange the above equation
a = (v - u) / t
= (0 m/s - 15.0 m/s) / (30.0 s)
= -0.5 m/s²
Now, use another equation of motion
S = ut + 0.5at²
S = (15.0 m/s × 30.0 s) + [0.5 × -0.5 m/s² × (30.0 s)^2]
= 225 m
∴ Train covers 225 m before it stops.
v = u + at
Rearrange the above equation
a = (v - u) / t
= (0 m/s - 15.0 m/s) / (30.0 s)
= -0.5 m/s²
Now, use another equation of motion
S = ut + 0.5at²
S = (15.0 m/s × 30.0 s) + [0.5 × -0.5 m/s² × (30.0 s)^2]
= 225 m
∴ Train covers 225 m before it stops.
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