Answer:
25 m/s
Explanation:
First we should define the variables
T=4
Dx = 100
ay=-9.8
ax=0
We can use formula 1 from the BIG 5
x=(v+v0)t/2
By plugging in our variables we can get 100=4(v+v0)/2
Which is 50=v+v0
v=v0 since horizontal acceleration always equals zero
so 2v0 = 50
v0 = 25
Answer:
3) Ep = 13243.5[J]
4) v = 17.15 [m/s]
Explanation:
3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.
m = mass = 90 [kg]
h = elevation = 15 [m]
Potential energy is defined as the product of mass by gravity by height.
![E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3Dm%2Ag%2Ah%5C%5CE_%7Bp%7D%3D90%2A9.81%2A15%5C%5CE_%7Bp%7D%3D13243.5%5BJ%5D)
This energy will be transformed into kinetic energy.
Ek = 13243.5 [J]
4) The velocity can be determined by defining the kinetic energy, as shown below.
![E_{k}=\frac{1}{2} *m*v^{2} \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%20%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B2%2AE_%7Bk%7D%20%7D%7Bm%7D%20%7D%5C%5C%20v%3D%20%5Csqrt%7B%5Cfrac%7B2%2A13243.5%20%7D%7B90%7D%20%7D%5C%5Cv%3D17.15%5Bm%2Fs%5D)
Acceleration x time = velocity
Since you're given acceleration and time, just plug the values into the equation.
3
x 1.1 s = ?
Solve that equation, and remember your velocity should be in m/s.
Answer:
From the narrative in the question, there seem to have been a break failure and the ordered step of response to this problem is to
1) Put on the hazard light to inform other road users of a problem or potential fault with your car and so they should continue their journey with caution.
2) Avoid pressing on the acceleration pedal as this might cause the car to gradually slow down due to friction and gravity
3)Try navigate the car to the service lane. This is the less busy lane where cars are sometimes parked briefly.
4) Continuously pump the breaks to try stop the car. Continuously pumping the breaks might just help you build enough pressure to stop the car because often time, there are some pressure left in the break.
5) At this point, the speed of the car should be relatively slow. So at this point, you could try apply the emergency hand break. Do not pull the emergency hand breaks if the car is on high speed. Doing this may cause the car to skid off the road.
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
