What Is the Total Displacement and Angle South of West for the Motorist's Trip?

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The motorist's trip involves three segments: traveling south at 24.0 m/s for 3 minutes, west at 25.0 m/s for 3 minutes, and northwest at 30.0 m/s for 1 minute. To calculate total displacement, it's crucial to convert the travel time from minutes to seconds for accurate results. The initial calculations for displacement from the first two segments should consider the correct vector addition and unit conversions. The final displacement vector is determined by combining these segments and calculating the resultant vector's magnitude and direction. Accurate unit handling is essential to avoid significant errors in the final answer.
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A motorist drives south at 24.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 3.00 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 7.00 min trip, find the following values.

a) total vector displacement and degrees south of west.

For the displacement I try to find the total displacement southwest of the first two vectors which should be sqrt(72^2+75^2) and get 103.966. From there I draw a line 30m northwest of it and at the endpoint of that line I draw another line to the origin. This 3rd line should be the hypotenuse of the triangle so I try adding 103.96^2 + 30^2 and taking the square root of it to get 107.28m but when I enter it on webassign it says I'm ORDERS OF MAGNITUDE WRONG? No idea what I'm doing wrong here and I don't think I can solve the second part without the first part.
 
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Welcome to PF.

Careful with your units.

Minutes are a little different than seconds.
 

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