A scooterist is approaching a circular turn of radius 80 m. He reduced his speed from 27 kmh−1 at constant rate of 0.5 ms−2. Find net acceleration and angle made by it wrt tangential deceleration
A
0.86 ms−2,54∘
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B
0.68 ms−2,45∘
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C
1.0 ms−2,45∘
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D
0.5 ms−2,45∘
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Solution
The correct option is A0.86 ms−2,54∘
Centripetal acceleration ar=v2r=(27×10003600)280 =7.5×7.580=0.703 ms−2 Net acceleration, a=√a2r+a2t where at = tangential deceleration =0.5ms−2
∴a=√0.7032+0.52=0.86ms−2 and θ=tan−1arat=tan−10.7030.5=54∘