A particle is moving with a constant acceleration. Its velocity is reduced to zero in 5 s and it covered a distance of 100 m in this direction. The distance covered by the particle in the next 5s is
Answers
Answer:
A = (v - u)/t
= (0 - u)/t
= -u / (5 s)
S = ut + 0.5at²
100 = 5u + (0.5 × [-u / (5 s)] × (5 s)²)
100 = 5u - 2.5u
100 = 2.5u
u = 40 m/s
If u = 40 m/s then
a = -u / (5 s)
= -(40 m/s) / (5 s)
= -8 m/s²
Displacement covered in 10 seconds
S = ut + 0.5at²
= (40 m/s × 10 s) + [0.5 × (-8 m/s² × (10 s)²]
= 0 m
Particle travelled 100 m in first 5 seconds and in next 5 seconds it travels 100 m back to initial position. Hence we got Displacement after 10 seconds equal to zero.
Explanation:
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Answer:
A = (v - u)/t
= (0 - u)/t
= -u / (5 s)
S = ut + 0.5at²
100 = 5u + (0.5 × [-u / (5 s)] × (5 s)²)
100 = 5u - 2.5u
100 = 2.5u
u = 40 m/s
If u = 40 m/s then
a = -u / (5 s)
= -(40 m/s) / (5 s)
= -8 m/s²
Displacement covered in 10 seconds
S = ut + 0.5at²
= (40 m/s × 10 s) + [0.5 × (-8 m/s² × (10 s)²]
= 0 m
Particle travelled 100 m in first 5 seconds and in next 5 seconds it travels 100 m back to initial position. Hence we got Displacement after 10 seconds equal to zero.