anonymous
  • anonymous
you drive with a constant speed of 13.5 m/s for 30 s. you then accelerate for 10s to a speed of 22 m/s you then slow to a stop in 10s how far have you traveled?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
case I when costt. v= 13.5 m/s time taken,t = 30 s dist. covered,\[s_{1}\]= v *t = 13.5 * 30 =405 m case II initial vel. , u = 13.5 m/s final vel. , v= 22 m/s time taken ,\[ t _{2}=10 s\] thus acc. ,a = (v-u)/t = (22- 13.5)/10 = .085 m/s^2 hence, \[v ^{2}-u^{2}=2as\] \[22^{2}-13.5^{2}= 2(0.85)*s\] 484-182.25 = 1.7 *s \[s _{2}=177.5 m\] caseIII v=0 u=22 m/s t=10 s a=(v-u)/t = -2.2m/s^2 \[v ^{2}-u ^{2}=2as\] -22 *22 /[2 *(-2.2)] =s \[s_{3}=110 m\] total dist.= \[s _{1}+s _{2}+s _{3}= \] 405 + 177.5 + 110 = 692.5 m
anonymous
  • anonymous
|dw:1444222868559:dw|
anonymous
  • anonymous
the area under the graph gives the distance covered . this is 2nd method to calculate the distance traveled. area = area(sq.) + ar(trapezium) + ar(triangle) u get the 3 equations in a single step

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