Hanky

A car travels 59.5 miles at 53.0° S of W. The car then makes a 90.0° turn to the left and travels an additional 90.3 miles. What is the car's final displacement? I have set up the diagram, and I believe I am supposed to use the pythagorean theorem. Well l...

A projectile is fired at 37.0° above the horizontal with an initial speed of 65.0 m/s. Find it's range. Well what I did was get the x and y components for the vector by: V-subX= (65 m/s) (cos(37°)) = 51.91 V-subY= (65 m/s) (sin(37°)) = 39.11 I'm not sure...

find an equation to the curve at the point corresponding to the given value of the parameter. x = tcost y = tsint when t = π i know I am supposed to find dy and dx which is: dy = (product form) t*-sint + 1*cost simplifying = -tsint+cost dx = tcost+sint no...

well, this was what I decided to go with. since it has designated that he made a 90° turn; We can indicate that it is a right triangle. So, I performed this: √( (59.5)^2 + (93.0)^2 )

Whoops sorry forgot to give my value after doing the calculation. My answer is: 110.4 miles, the miles squared gets square rooted so then it becomes just miles, right?

Ok, so I am guessing I need to use the angle given to me. Now I have to find the hypotenuse length and angle? I think?

if I did it my way, would I be taking steps to solve the problem?

or actually that is what you said, sorry.

Ok thank you! I accidentally miscalculated my hypotenuse... so now my final answer is: 108.14 miles 56.6°East of South, because it passes the third quadrant by 19.6° which is why you subtracted 90°

Steve, thank you for replying. I'm sorry but I'm not exactly sure why you used the function y(t). also I don't know if that's a specialized equation. We are only supposed to use the 4 kinematic equations.