Asked by Amy
                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.
            
        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.
Answers
                    Answered by
            Damon
            
    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
    
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
                    Answered by
            Damon
            
    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
                    Answered by
            Damon
            
    Oh and the magnitude of the velocity in #4 is sqrt (1^2 + 5^2)
the hypotenuse
    
the hypotenuse
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.