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Step 4 Alternatively, we can compute the acceleration by first finding its x- and y-components. For the x-component of acceleration, we have 245 245 N 1 kg - m/s? 1N 0.75 0.754 m/s2 325 kg and, for the y-component of acceleration, 495 495 N 1 kg - m/s? 1N ΣF ay = - 1.52 1.52 m/s?. m 325 kg Step 5 The resulting acceleration is computed from its components. For its magnitude, we have 2.3 x m²/s = 1.69 m/s? %3D

Step 4 Alternatively, we can compute the acceleration by first finding its x- and y-components. For the x-component of acceleration, we have 245 245 N 1 kg - m/s? 1N 0.75 0.754 m/s2 325 kg and, for the y-component of acceleration, 495 495 N 1 kg - m/s? 1N ΣF ay = - 1.52 1.52 m/s?. m 325 kg Step 5 The resulting acceleration is computed from its components. For its magnitude, we have 2.3 x m²/s = 1.69 m/s? %3D

College Physics
College Physics
10th Edition
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter3: Vectors And Two-Dimensional Motion
Section: Chapter Questions
Problem 8P: A force F1, of magnitude 6.00 units acts on an object at the origin in a direction = 30.0 above the...
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Need help with STEP 5. I'm not sure why I got the answer wrong (marked in red X).

Step 3 The magnitude of the resulting acceleration of the boat is 552 552 N) 1 kg m/s? 1.7 m/s?. 1.698 1N 325 kg The acceleration is in the same direction of the force; that is a, F/m, so %3! R a, = 1.698 1.7 m/s? at 63.66 63.7 ° north of east. R Step 4 Alternatively, we can compute the acceleration by first finding its x- and y-components. For the x-component of acceleration, we have 245 245 N 1 kg · m/s? 1N ΣF. 0.75 0.754 m/s? 325 kg and, for the y-component of acceleration, 495 495 N EF 1 kg · m/s? 1.52 1.52 m/s?. 1N 325 kg Step 5 The resulting acceleration is computed from its components. For its magnitude, we have a 2.3 × m²/s = 1.69 m/s2 aR = +a and for its direction, 1.52 m/s2 e = tan = tan = 63.73 0.75 m/s?
Transcribed Image Text:Step 3 The magnitude of the resulting acceleration of the boat is 552 552 N) 1 kg m/s? 1.7 m/s?. 1.698 1N 325 kg The acceleration is in the same direction of the force; that is a, F/m, so %3! R a, = 1.698 1.7 m/s? at 63.66 63.7 ° north of east. R Step 4 Alternatively, we can compute the acceleration by first finding its x- and y-components. For the x-component of acceleration, we have 245 245 N 1 kg · m/s? 1N ΣF. 0.75 0.754 m/s? 325 kg and, for the y-component of acceleration, 495 495 N EF 1 kg · m/s? 1.52 1.52 m/s?. 1N 325 kg Step 5 The resulting acceleration is computed from its components. For its magnitude, we have a 2.3 × m²/s = 1.69 m/s2 aR = +a and for its direction, 1.52 m/s2 e = tan = tan = 63.73 0.75 m/s?
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