Answer:
78 km/h
Explanation:
If I normally drive a 12 hour trip at an average speed of 100 km/h, my destination has a total distance of:
- 100 km/h · 12 h = 1,200 km
Today, I drive the first 2/3 of the distance at 116 km/h. Let's first calculate what 2/3 of the normal distance is.
I've driven 800 km already. I need to drive 400 km more to reach my final destination. I need to figure out my average speed during this last 1/3 of the distance.
To do this, I first need to calculate how much time I spent driving 116 km/h for the past 800 km.
- 116 km/1 h = 800 km/? h
- 800 = 116 · ?
- ? = 800/116
- ? = 6.89655172
I spent 6.89655172 hours driving during the first 2/3 of the distance.
Now, I need to subtract this value from 12 hours to find the remaining time I have left.
- 12 h - 6.89655172 h = 5.10344828 h
Using this remaining time and my remaining distance, I can calculate my average speed.
- ? km/1 hr = 400 km/5.10344828 h
- 5.10344828 · ? = 400
- ? = 400/5.10344828
- ? = 78.3783783148
My average speed during the last third of the distance is around 78 km/h.
Answer:
it takes the car 4.362 seconds to cover the distance of 88.4 m.
Explanation:
The distance the car covers is given by the function
,
where 
, and 
, putting these in we get:
Now, when the car has moved to 88.4m, 
, or
which is a quadratic equation with solutions
We take the first solution 
, <em>since at that time the car is still moving right and decelerating</em>. The second solution 
describes the situation where the car has stopped decelerating and is now moving leftwards because the decelerating is leftwards, <em>which is utterly wrong because we know that cars do not start moving backwards after the brakes have stopped them! </em>
Thus, it takes the car 4.362 seconds to cover the distance of 88.4 m.
The answer is B as all the other options contain quantities not related to describing motion
Answer:
x = 4 m
Explanation:
For this exercise we must use the rotational equilibrium relationship, where we place zero at the turning point and counterclockwise rotations we will consider positive
as it indicates that the bar is in equilibrium, its center of mass coincides with the turning point, so the distance is zero and does not create torque on the system
∑τ = 0
W 3 - w x = 0
x = 3W / w
x = 3 Mg / mg
x = 3 M / m
let's calculate
x = 3 60/45
x = 4 m
