To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From them we will consider speed as the distance traveled per unit of time. Said unit of time will be cleared to find the total time taken to travel the given distance. Later with the calculated average times and distances, we will obtain the average speed.
PART A)
The time taken to travel a distance of 250km with a speed of 95km/h is



Time taken for the lunch is

The time taken travel a distance of 250km with a speed of 55km/h



The total time taken is



The average speed is the ratio of total distance and total time


PART B)
As the displacement is zero the average velocity is zero.
Answer:
F = - 59.375 N
Explanation:
GIVEN DATA:
Initial velocity = 11 m/s
final velocity = 1.5 m/s
let force be F
work done = mass* F = 4*F
we know that
Change in kinetic energy = work done
kinetic energy = 
kinetic energy =
= -237.5 kg m/s2
-237.5 = 4*F
F = - 59.375 N
The half-life equation
in which <em>n </em>is equal to the number of half-lives that have passed can be altered to solve for <em>n.</em>
<em>
</em>
<em>
</em>
Then, the number of half-lives that passed can be multiplied by the length of a half-life to find the total time.
<em>2 * 5700 = </em>11400 yr
Explanation:
It is given that,
Mass of the car 1, 
Initial speed of car 1,
(east)
Mass of the car 2, 
Initial speed of car 2,
(north)
(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :




The magnitude of speed,

V = 12.22 m/s
(b) Let
is the direction the wreckage move just after the collision. It is given by :



Hence, this is the required solution.
First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg.
F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N
Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m
Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356 N
Ratio = 0.356 N/589.18 N
<em>Ratio = 6.04</em>