Draw a vector diagram. The net force on particle 1 = F12 + F13 + F14 These forces have to be added as vectors.
We will resolve our forces along the direction 1-4 F12 (tot) = -kQq / a^2 in the direction of particle 4 F12 = -kQq *sin (45) / a^2 F12 = -kQq /( a^2 * sqrt(2) )
By symetry this is the same as F13 F13 = -kQq /( a^2 * sqrt(2) )
F14 = -kQQ / (Sqrt(2)*a) ^ 2
For net force on particle 1 :
F12+F13+F14 = 0 -2kQq /( a^2 * sqrt(2) ) + -kQQ / (Sqrt(2)*a) ^ 2 = 0
Some simple manipulation should give you :
Q/q = -2 sqrt(2)
The best answer would be a "non-directional hypothesis"
<span>This problem is solved by the equation of motion:
x = x0 + v0*t + 1/2*a*t^2,
Here x0 = 0, v0 = 40ft/sec and a = -5 ft/s^2, we need to solve for t:
v = v0 + a*t, solve how long does it take to stop: 0 = v0 + a*t --> a*t = -v0 --> t = -v0/a
-- > 40/5 = 8 seconds to stop.
In this time, the car travels x = 0 + 40*8 + 0.5*-5*8^2 ft ~ 160 ft.
Answer: The car travels 160 ft.</span>
So, the acceleration of the bicycle is approximately <u>-1.67 m/s²</u> or it can be said to be decelerating approximately <u>1.67 m/s²</u>.
<h3>Introduction</h3>
Hi ! Here I will help material about linear motion changes regularly, which is where you will hear a lot of the term acceleration. Acceleration occurs when an object's speed increases in a certain time interval. Acceleration can be negative which is called deceleration. The relationship between acceleration with velocity and time is manifested in the equation:

With the following conditions :
- a = acceleration (m/s²)
= object's final velocity (m/s)
= object's initial velocity (m/s)- t = interval of the time (s)
<h3>Problem Solving </h3>
We know that :
= object's final velocity = 4 m/s
= object's initial velocity = 12 m/s- t = interval of the time = 4.8 s
What was asked :
- a = acceleration = ... m/s²
Step by step :




So, the acceleration of the bicycle is about -1.67 m/s² or it can be said to be decelerating around 1.67 m/s².
Both vehicles experience the same change in momentum.
When the compact car rear ends a truck, they exert equal forces on each other, in accordance with Newton's third law. The force exerted by the car on the truck is equal to the force exerted by the truck on the car.The force acting on the two bodies act for the same interval of time.
According to Newton's second law, the force acting on a body is equal to the rate of change of momentum.
Since the forces are equal,

therefore, the change in momentum of the car
is equal to the change in the momentum of the truck
.
Thus, both the car and the truck experience the same change in their momentum. However due to the smaller mass of the car, the change in its velocity is greater than the change in the truck's velocity, which has a larger mass.