A train starts from a station and moves with a uniform acceleration for 2 minutes. If it covers a distance of 400m in this period. Find the acceleration and final velocity attained.
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Use equation of motion
S = ut + 0.5at²
Initial velocity (u) is zero
So,
a = 2S / t²
= (2 × 400 m) / (2 × 60 s)²
= 5.55 × 10⁻⁴ m/s²
Acceleration is 5.55 × 10⁻⁴ m/s²
v = u + at
= 0 + (5.55 × 10⁻⁴ m/s² × 2 × 60 s)
≈ 0.067 m/s
Final velocity is approximately 0.067 m/s
S = ut + 0.5at²
Initial velocity (u) is zero
So,
a = 2S / t²
= (2 × 400 m) / (2 × 60 s)²
= 5.55 × 10⁻⁴ m/s²
Acceleration is 5.55 × 10⁻⁴ m/s²
v = u + at
= 0 + (5.55 × 10⁻⁴ m/s² × 2 × 60 s)
≈ 0.067 m/s
Final velocity is approximately 0.067 m/s
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