Answer:
1.) Time t = 3.1 seconds
2.) Height h = 46 metres
Explanation:
given that the initial velocity U = 30 m/s
At the top of the trajectory, the final velocity V = 0
Using first equation of motion
V = U - gt
g is negative 9.81m/^2 as the object is going against the gravity.
Substitute all the parameters into the formula
0 = 30 - 9.81t
9.81t = 30
Make t the subject of formula
t = 30/9.81
t = 3.058 seconds
t = 3.1 seconds approximately
Therefore, it will take 3.1 seconds to reach to reach the top of its trajectory.
2.) The height it will go can be calculated by using second equation of motion
h = ut - 1/2gt^2
Substitutes U, g and t into the formula
h = 30(3.1) - 1/2 × 9.8 × 3.1^2
h = 93 - 47.089
h = 45.911 m
It will go 46 metres approximately high.
So we want to know what is the magnitude of the horizontal component of acceleration ah if we know that the overall acceleration a=12 m/s^2 and the angle of overall acceleration and the horizontal acceleration is α=50°. We know that ah=a*cosα. So now it isn't hard to get the horizontal component: ah=12*cos50=12*0.64=7.71 m/s^2. So the correct answer is ah=7.71 m/s^2.
Answer:
Ok, so the car has traversed a total of 2 + 4 + 2 + 6 = 14 miles.
That's the car's distance! :)
For the displacement, you can draw out a diagram. I'll try my best to make one using a keyboard xDDD
- - - - |
| |
| - - - start - end!
So, this is the journey of the car: Going two up, 4 left, 2 down, and 6 right.
This ends up 2 to the right of the beginning!
That's a 2 mile displacement :)
Can you give me brainliest for my splendid diagram? xD
