angle between the velocity a bar =2i cap +2j cap -2k cap and b bar= 3i cap- 4j cap
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Given :- A car initially at rest, picks up a velocity of 72 km/h in\:\dfrac{1}{4}\:41 minute To Calculate :- Acceleration Distance covered by the car Solution :- We have, Initial Velocity(u) of the car = 0 m/s Final Velocity(v) of the car = 20 m/s Time taken(t) = \:\dfrac{1}{4}\:41 minute = 15 s We know that, \boxed{\sf v = u + at} \: \: (\bf 1st \: equation \: of \: motion)v=u+at(1stequationofmotion) Substituting the values we get, :\implies\sf20 = 0 + a \times 15:⟹20=0+a×15 :\implies\sf20 = 15a:⟹20=15a :\implies\sf\cancel- 15a = \cancel- 20:⟹−15a=−20 :\implies\sf a = \dfrac{\cancel{20}}{\cancel{15}}:⟹a=1520 :\implies\sf\underline{\boxed{\blue{\sf a = 1.3\:m/s^{2}}}}:⟹a=1.3m/s2 \green{\therefore\sf Acceleration \: is \: 1.3\:m/s^{2}}∴Accelerationis1.3m/s2 We know that, \boxed{\sf s = ut + \dfrac{1}{2}at^{2}} \: \: (\bf 2nd \: equation \: of \: motion)s=ut+21at2(2ndequationofmotion) Substituting the values we get, :\implies\sf s = 0 \times 15 + \dfrac{1}{2} \times 1.3 \times 15^{2}:⟹s=0×15+21×1.3×152 :\implies\sf s = 0 \times 15 + \dfrac{1}{2} \times 1.3 \times 225:⟹s=0×15+21×1.3×225 :\implies\sf s = 0 \times 15 + \dfrac{1}{\cancel2} \times \cancel{292.5}:⟹s=0×15+21×292.5 :\implies\sf s = 0 \times 15 + 146.25:⟹s=0×15+146.25 :\implies\sf s = 0 + 146.25:⟹s=0+146.25 :\implies\sf\underline{\boxed{\blue{\sf s = 146.25\:m}}}:⟹s=146.25m \green{\therefore\sf Distance \: travelled \: is \: 146.25\:m}∴Distancetravelledis146.25m _____________________________ Additional Information :- ★ 3rd equation of motion :- \boxed{\sf v^{2} - u^{2} = 2as}v2−u2=2as Where, v = Final Velocity u = Initial Velocity a = Acceleration s = Distance travelled ______________________________