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) 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
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
Oh and the magnitude of the velocity in #4 is sqrt (1^2 + 5^2)
the hypotenuse
To find the resultant velocity of the boat, we need to consider both the velocity of the boat and the velocity of the river. Since the boat is moving eastward and the river is flowing southward, these velocities are perpendicular to each other.
To find the resultant velocity, we can use the concept of vector addition. We can calculate the resultant velocity by finding the vector sum of the boat's velocity and the river's velocity.
Let's break down the velocities into their horizontal and vertical components:
1) Boat's velocity: 5 m/s eastward
- Horizontal component (x-direction): 5 m/s
- Vertical component (y-direction): 0 m/s (since the boat is not moving north or south)
2) River's velocity: 1 m/s southward
- Horizontal component (x-direction): 0 m/s (since the river is not flowing east or west)
- Vertical component (y-direction): -1 m/s (negative sign indicates the direction opposite to the boat's motion)
Now, we can add the horizontal and vertical components separately to find the resultant velocity:
Horizontal component: 5 m/s + 0 m/s = 5 m/s
Vertical component: 0 m/s + (-1 m/s) = -1 m/s
The resultant velocity can be found by combining the horizontal and vertical components. Using the Pythagorean theorem:
Resultant velocity = √(horizontal component)^2 + (vertical component)^2
Resultant velocity = √((5 m/s)^2 + (-1 m/s)^2)
Resultant velocity = √(25 m^2/s^2 + 1 m^2/s^2)
Resultant velocity = √(26 m^2/s^2)
Resultant velocity = √26 m/s
So, the resultant velocity of the boat is approximately 5.1 m/s, in a direction between east and southeast.