Answer:
<em>The distance the car traveled is 21.45 m</em>
Explanation:
<u>Motion With Constant Acceleration
</u>
It occurs when an object changes its velocity at the same rate thus the acceleration is constant.
The relation between the initial and final speeds is:
![v_f=v_o+at\qquad\qquad [1]](https://tex.z-dn.net/?f=v_f%3Dv_o%2Bat%5Cqquad%5Cqquad%20%5B1%5D)
Where:
a = acceleration
vo = initial speed
vf = final speed
t = time
The distance traveled by the object is given by:
![\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3Dv_o.t%2B%5Cfrac%7Ba.t%5E2%7D%7B2%7D%5Cqquad%5Cqquad%20%5B2%5D)
Solving [1] for a:
Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:
The distance is now calculated with [2]:
x = 21.45 m
The distance the car traveled is 21.45 m
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
The answer is either C or D.
