Vehicle weight shifts can be backward forward. In this particular case accelerating to fast would cause a shift of weight backwards. Breaking too quickly on the other hand would cause weight to shift forward. You have seen this while in a car or a bus at the traffic light. As the vehicle breaks you are pulled forward as it starts moving you are pulled backwards.
Answer: 0.25 m/s
Explanation: Speed = wavelengt · frequency
v = λf and frequency is 1/period f = 1/T
Then v = λ/T = 5 m / 20 s = 0.25 m/s
Answer:
4.18
Explanation:
Givens
The car's initial velocity
= 0 and covering a distance Δx = 1/4 mi = 402.336 m in a time interval t = 4.43 s.
Knowns
We know that the maximum static friction force is given by:
μ_s*n (1)
Where μ_s is the coefficient of static friction and n is the normal force.
Calculations
(a) First, we calculate the acceleration needed to achieve this goal by substituting the given values into a proper kinematic equation as follows:
Δx=
a=41 m/s
This is the acceleration provided by the engine. Applying Newton's second law on the car, so in equilibrium, when the car is about to move, we find that:

Substituting (3) into (1), we get:
μ_s*m*g
Equating this equation with (4), we get:
ma= μ_s*m*g
μ_s=a/g
=4.18
Salt is the solute in that situation
initial speed of the racer is given as


after applied force the final speed is given as


now during this speed change the racer will cover total distance 185 m
so here we will use kinematics



now the force that chute will exert on the racer will be given as



B) here following is the strategy for solving it
1. first we used kinematics to find the acceleration of the car
2. then we used Newton's II law (F = ma) to find the force