A) 2 m/s
B) 4 m/s
C) 6 m/s
D) 8 m/s
Correct Answer: B
Solution :Let the velocity at \[t=2.5\,s\]be v. So the change in the velocity from\[t=0\]to \[t=2.5\,s\]is also v. Change in velocity = Area under a -t curve with proper sign. or,\[v=4\times 1.5-2\times 1\] = 4 m/s. Hence, the correction option is (b).
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