Question

A car driving on the turnpike accelerates uniformly in the positive x direction from 97.0 ft/s (66.1 mph ) to 127 ft/s (86.6 mph ) in 3.20 s .

How much time does it take the car to overtake the truck?

How far was the car behind the truck initially?

What is the speed of the truck when they are abreast?

What is the speed of the car when they are abreast?

Answer 1

a turnpike is a toll-gate i reckon

Answer 2

And this is a statement rather than a question


\(\color{#0cbb34}{\text{Originally Posted by}}\) @BlackReaper
A car driving on the turnpike accelerates uniformly in the positive x direction from 97.0 ft/s (66.1 mph ) to 127 ft/s (86.6 mph ) in 3.20 s .
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 3

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth’'s surface and is to reach a maximum height of 960 m above the earth'’s surface. The rocket’'s engines give the rocket an upward acceleration of 16.0 m/s2 during the time T that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance.

What must be the value of T in order for the rocket to reach the required altitude?

Answer 4

Assuming ground level to be at zero potential.

the net acceleration of rocket for time interval 'T' =a= (acceleration due to engine - acceleration due to gravity)=16-10=6 m/s^2

Height achieved in time interval 'T'=h=\(\frac{1}{2} 6 T^2\)
speed of rocket after time 'T'=6T

Answer 5

Now u may apply law of mechanical conservation of energy from the point where the engine shuts off to the highest point where speed becomes zero.

\(KE _i +PE_i=KE_f+PE_f\)
\[\frac{1}{2}m (6T)^2+mgh=mgh_{\max}\]h=\(3T^2\)
h_{max}=960 m

so\[\frac{1}{2} \cancel m (6T)^2+\cancel m g (3T^2)= \cancel m g (960)\]solve for 'T'