A) 15 m/s
B) 25 m/s
C) 35 m/s
D) 45 m/s
Correct Answer: A
Solution :
Acceleration \[a=\frac{dv}{dt}=2(t-1)\] Integrating, \[v={{t}^{2}}-2t+c\] Velocity\[v=0\]at \[t=0\] \[\therefore \] \[c=0\] \[v={{t}^{2}}-2t\] \[\therefore \] Velocity at 5 s \[v={{(5)}^{2}}-2\times 5\] \[=15m/s\]You need to login to perform this action.
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