Answer:
d. 6.0 m
Explanation:
Given;
initial velocity of the car, u = 7.0 m/s
distance traveled by the car, d = 1.5 m
Assuming the car to be decelerating at a constant rate when the brakes were applied;
v² = u² + 2(-a)s
v² = u² - 2as
where;
v is the final velocity of the car when it stops
0 = u² - 2as
2as = u²
a = u² / 2s
a = (7)² / (2 x 1.5)
a = 16.333 m/s
When the velocity is 14 m/s
v² = u² - 2as
0 = u² - 2as
2as = u²
s = u² / 2a
s = (14)² / (2 x 16.333)
s = 6.0 m
Therefore, If the car had been moving at 14 m/s, it would have traveled 6.0 m before stopping.
The correct option is d
You would be correct.
Because you have only JUST released the arrow, and how close he is to the target, it would have the same amount of energy when it strikes the target. Yes, the kinetic energy would be destroyed when you hit the target but not right away. And yes, the potential energy would also be destroyed once you release the arrow, but it goes straight back once it stops moving, aka when it hits the target, although it has only just stopped moving.
Hope this helps!
Answer:
The final velocity of the object is, 
= 27 m/s
Explanation:
Given,
The acceleration of the object, a = 1000 m/s²
The initial displacement of the object, 
= 0 m
The final displacement of the object, 
= 0.75 m
The initial velocity of the object will be, 
= o m/s
The final velocity of the object, = ?
The average velocity of the object,
v = ( - )/ t
= 0.75 / t
The acceleration is given by the relation
a = v / t
1000 m/s² = 0.75 / t²
t² = 7.5 x 10⁻⁴
t = 0.027 s
Using the I equation of motion,
= u + at
Substituting the values
= 0 + 1000 x 0.027
= 27 m/s
Hence, the final velocity of the object is, = 27 m/s
