Question

2. A particle is pushed along a horizontal surface in such a way that it starts with a velocity of 14
m/s. Its velocity decreases at a rate of 0.6 m/s^2, the distance covered by it before coming to
rest is
​

Answer 1

Answer:

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Explanation:

u = 14 m/s

v = 0 m/s

a = - 0.6 m/s²

v² - u² = 2as

0 - 196 = -2*0.6*s

196 = 1.2s

s = 196/1.2 = 163.33 m

Answer 2

Given:-

• Initial velocity (u) = 14m/s

• Final Velocity (v) = 0m/s

• Acceleration (a) = -0.6m/s²

To Find:-

• Distance covered by particle (s)

Solution:-

By using 3rd equation of motion

→ v² = u²+2as

Substitute the value we get

→ 0² = 14² + 2×(-0.6)×s

→ 0 = 196 + (-1.2)×s

→ -196 = -1.2×s

→ s = -196/-1.2

→ s = 196/1.2

→ s = 163.33 m

∴ The distance covered by the particle is 163.33 metre.