Question

A bus starting from rest moves with a uniform acceleration of 0.1m/s2 for 2 minutes .Find (a)the speed acquired (b) the distance travelled.

Answer 1

u=0,

a=0.1m/s²,

t=2min=120sec,

v=?;

(a). a=(v-u)/t

0.1=(v-0)/120

0.1×120=v

v=12m/s

(b). v²-u²=2aS

12²-0²=2×0.1×S

S=(12×12)/(2×0.1)

S=60×12

S=720m

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Answer 2

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