A) zero
B) \[\frac{1}{\sqrt{2}}m/{{s}^{2}}\]towards north-west
C) \[\frac{1}{\sqrt{2}}m/{{s}^{2}}\]towards north-east
D) \[\frac{1}{2}m/{{s}^{2}}\]towards north-west
E) \[\frac{1}{2}m/{{s}^{2}}\]towards north
Correct Answer: B
Solution :
Average acceleration \[=\frac{\Delta \overrightarrow{v}}{\Delta t}\] \[\Delta v=\sqrt{v_{1}^{2}+v_{2}^{2}}=\sqrt{{{5}^{2}}+{{5}^{2}}}=5\sqrt{2}\] \[\therefore \]Average acceleration \[=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}}m/{{s}^{2}}\]. towards north-west.You need to login to perform this action.
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