A) 1
B) 3
C) 4
D) 5
Correct Answer: C
Solution :
\[s=2{{t}^{2}}-3t+1;\,\,\,\therefore \frac{ds}{dt}=v=4t-3\] At \[t=0{{\left( \frac{ds}{dt} \right)}_{t.0}}=-3={{v}_{1}}\] Now, at \[t=1\], we get \[{{\left( \frac{ds}{dt} \right)}_{t=1}}=4-3=1={{v}_{2}}\] Hence rate of velocity \[={{v}_{2}}-{{v}_{1}}=1-(-3)=4\] Aliter : Given that \[s=2{{t}^{2}}-3t+1\] \[\frac{ds}{dt}=4t-3\](velocity), Again \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=4\](acceleration).
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