1) right
2) v= a t
41 = a (8)
a = 5.125
x = 0 + 0 + .5 (5.125)(8)^2
x = 164 meters
I think you maybe forgot the 1/2 in (1/2) a t^2
3)Vfinal = -15
Vinitial = 35
Vfinal - Vinitial = -15 -35 = -50km/hr
4) I assume you mean the boat is HEADING east, NOT GOING east. To go east the pilot must steer somewhat upstream and I do not think that is what the teacher probably means
If it is HEADED east it is going 5 m/s east and 1 m/s south
the tangent of the angle south of east is 1/5
so the angle south of east is tan^-1 (.20)
which is 11.3 degrees south of east is the course made good.
On a nautical chart which calls north zero and east 90 it would be 90 +11.3 = 101 degrees True
Can someone please check my answers for the first three questions and help me with the last one?
1)A racecar accelerates from rest at +7.2 m/s2 for 4.1 seconds. How fast will it be going at the end of that time?
Answer: 29.5 m/s
2)A racecar starts from rest and is accelerated uniformly to +41 m/s in 8.0 s. What will be the car's displacement?
Answer: 328 m
3)A rubber ball strikes a wall at a velocity of +35 km/hr and rebounds with a velocity of -15 km/hr. The change in velociy is:
Answer: +50 km/hr
I need some help with this last question...
A boat going 5 m/s eastward is crossing a river that flows southward at 1 m/s. Find the resultant velocity of the boat.
3 answers
Oh another way to do number two is to say if it accelerated uniformly from 0 to 41, the average speed during the eight seconds would be 20.5 for the eight seconds.
20.5 m/s * 8 s = 164 meters
20.5 m/s * 8 s = 164 meters
Oh and the magnitude of the velocity in #4 is sqrt (1^2 + 5^2)
the hypotenuse
the hypotenuse