A) \[6\text{ }m/{{s}^{2}},500\text{ }m\]
B) \[3\text{ }m/{{s}^{2}},200m\]
C) \[-6\text{ }m/{{s}^{2}},500m\]
D) \[-3\text{ }m/{{s}^{2}},500m\]
Correct Answer: A
Solution :
Rate of change of velocity is defined as acceleration. \[\therefore \] \[a=\frac{\Delta V}{\Delta t}\] \[\Delta t=20-0=20\text{ }s,\text{ }\Delta V=20\text{ }m/s\] \[\therefore \] \[{{a}_{1}}=\frac{20}{20}=1\,m/{{s}^{2}}\] Also, acceleration from 30 s to 40 s is \[{{a}_{2}}=\frac{80-20}{40-30}=\frac{60}{10}=6\,m/{{s}^{2}}\] \[Distance=velocity\times time\] Distance = area of trapezium Distance \[=\frac{1}{2}[(20+80)(40-30)]\text{ }m\] Distance\[=\frac{100\times 10}{2}=500\,m\]
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