A bus starting from rest moves with a uniform acceleration of 0.1m/s(square) for 2min. .Find (a) the speed acquired,(d) the distance travelled.
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Answer:
Given
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120s
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120sTo Find
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120sTo FindDistance travelled
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120sTo FindDistance travelledSolution
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120sTo FindDistance travelledSolutionAs we know that
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120sTo FindDistance travelledSolutionAs we know that→ s = ut + ½ at² {2nd equation of motion }
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120sTo FindDistance travelledSolutionAs we know that→ s = ut + ½ at² {2nd equation of motion }→ s = 0 × 120 + ½ × 0.1 × 120 × 120
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120sTo FindDistance travelledSolutionAs we know that→ s = ut + ½ at² {2nd equation of motion }→ s = 0 × 120 + ½ × 0.1 × 120 × 120→ s = 0.1 × 60 × 120
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120sTo FindDistance travelledSolutionAs we know that→ s = ut + ½ at² {2nd equation of motion }→ s = 0 × 120 + ½ × 0.1 × 120 × 120→ s = 0.1 × 60 × 120→ s = 720m
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120sTo FindDistance travelledSolutionAs we know that→ s = ut + ½ at² {2nd equation of motion }→ s = 0 × 120 + ½ × 0.1 × 120 × 120→ s = 0.1 × 60 × 120→ s = 720mHence,
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120sTo FindDistance travelledSolutionAs we know that→ s = ut + ½ at² {2nd equation of motion }→ s = 0 × 120 + ½ × 0.1 × 120 × 120→ s = 0.1 × 60 × 120→ s = 720mHence,Distance travelled by bus is720m
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120sTo FindDistance travelledSolutionAs we know that→ s = ut + ½ at² {2nd equation of motion }→ s = 0 × 120 + ½ × 0.1 × 120 × 120→ s = 0.1 × 60 × 120→ s = 720mHence,Distance travelled by bus is720mExtra Information
GivenInitial velocity (u) = 0 {rest position}Acceleration (a) = 0.1m/s²time taken (t) = 2min = 2 × 60 = 120sTo FindDistance travelledSolutionAs we know that→ s = ut + ½ at² {2nd equation of motion }→ s = 0 × 120 + ½ × 0.1 × 120 × 120→ s = 0.1 × 60 × 120→ s = 720mHence,Distance travelled by bus is720mExtra InformationForce = mass × accelerationv = u + at {First equation of motion}v² = u² + 2as {third equation of motion}Pressure = Force/Area