A bus decreases its speed from 80 km/h to 60 km/h in 5 s. The acceleration of the bus is

  1. 1.11 m/s2 acceleration
  2. 1.11 m/s2 deacceleration
  3. 2.22 m/s2 acceleration
  4. 2.22 m/s2 deacceleration

Answer (Detailed Solution Below)

Option 2 : 1.11 m/s2 deacceleration
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Detailed Solution

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CONCEPT:

  • Equation of Kinematics: These are the various relations between u, v, a, t and s for the particle moving with uniform acceleration where the notations are used as:
  • Equations of motion can be written as

V = U + at

\(s =ut+\frac{1}{2}{at^{2}}\)

V2 =U2+ 2as

Where u = Initial velocity of the particle at time t = 0 sec

v = Final velocity at time t sec

a = Acceleration of the particle

s = Distance travelled in time t sec

EXPLANATION:

Given - Initial velocity (u) = 80 km/h = 22.22 m/s, Final velocity (v) = 60 km/h = 16.67 m/s, and time (t) = 5 s.

  • As per first law of motion

⇒ v = u + at

⇒ 16.67 = 22.22 + (a × 5)

\(\Rightarrow a = \frac{{16.67 - 22.22}}{5} = - \frac{{5.55}}{5} = - 1.11\;\, m/s^2\)

  • The negative sign shows that the bus is deaccelerating at a rate of 1.11 m/s2.
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