How much time will it take for a bug to travel 5 meters across the floor if it is traveling at 1 m/s?

Physics

1 answer:

Karolina [17]3 years ago

5 0

If it's traveling at 1 m/s, it takes five seconds to travel five meters.

You might be interested in

I don't understand help

Bumek [7]

Answer:

Pe is stored energy to do work Pe is stored in objects that are at rest

KE defined as the energy of motion

began to move

Not sure about the rest

hope this helps

Explanation:

6 0

3 years ago

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

Thepotemich [5.8K]

Answer:

$T=6.75s$

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on and the second when the rocket's engines is off. So, we need to know the height ( $h_1$ ) that the rocket reaches while its engine is on and we need to know the distance ( $h_2$ ) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

$h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\$

and the final speed is:

$v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T$

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

$v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2$

The sum of these heights will give us the total height, which is known:

$h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s$