Wow! 2.95m/s is a mighty fast pace for a backpacker. Must have one of those Star Wars anti-gravity packs. Also, I would be curious as to why she passed her destination and then walked back.
<span>Anyway, it goes like this: </span>
<span>Say the time walking east is 't', and the total time is 'T'. </span>
<span>Then 5340 m + .511 t = 1.43 T </span>
<span>(This assumes that velocity is positive in both directions) </span>
<span>Two unknowns in one equation. But you also know that the time spent walking west is </span>
<span>5340m/ 2.95m/s = 1810 sec. </span>
<span>and the total time T = 1810 +t </span>
<span>Substitute this into the first equation, and you can solve for t = 3092 sec. </span>
<span>Then T = 4902 sec. and distance walked east is .511t = 1580m.</span>
Answer:
315 mi is the answer.
Explanation:
speed (s) = 70mph
time (t) = 4.5 hours
distance (d) = speed × time
d = 70 × 4.5
d = 315
∴ distance will be 315 mi.
<span>The answer is 8.8 m/s^2. The acceleration (a) is the change in velocity (v) in time (t): a=Δv/t. Δv=v2-v1. Since the car starts from the rest, v1=0 m/s. The final velocity (v2) can be calculated from kinetic energy (KE) and the mass (m) of the car since KE=1/2*m*v2^2. From here: v2^2=2KE/m. v2=√(2KE/m). KE=3.13×10^5 J=3.13×10^5 kg*m^2/s^2, m=290 kg. v2=√(2*3.13×10^5 kg*m^2/s^2/290 kg)=√(2,158.62 m^2/s^2)=46.5 m/s. So, Δv=v2-v1=46.5 m/s - 0 m/s= 46.5 m/s. Now, we have Δv = 46.5 m/s and t=5.3 s, so the acceleration is: a=Δv/t=46.5 m/s/5.3 s = 8.8 m/s^2.</span>
(a) The magnitude of the wind as it is measured on the boat will be the result of the two vectors. Since they are at 90°, the resultant can be determined by the Pythagorean theorem.
R = sqrt ((20 knots)² + (17 knots)²)
R = sqrt (400 + 289)
R = 26.24 knots
The direction of the wind will have to be angle between the boat and the resultant.
cos θ = (20 knots)/(26.24 knots)
θ = 40.36°
Hence, the direction is 40.36° east of north.
(b) As stated, the wind is blowing in the direction that is to the east. This means that it only has one direction. Parallel to the motion of the boat, the magnitude of the wind velocity will have to be zero.
