Acar driving at an initial speed of 10.0 m/s accelerates on a straight road at 3.00 m/s^2. a) what is the speed of the car after one quarter of a mile? (1 mile =1.609km). b) the driver in part a slams on the brakes after reaching the quarter mile. if the car can decelerate at a rate of 4.50 m/s^2, what is the stopping distance of the car?

The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.

Explanation:

Given that,

Initial speed = 10.0 m/s

Acceleration = 3.00 m/s^2

Distance

We need to calculate the speed of the car,

Using equation of motion

Where, u = initial velocity

v = final velocity

a = acceleration

Put the value in the equation

(B). We need to calculate the stopping distance of the car,

Using equation of motion again

Here,initial velocity = 50.13 m/s

Final velocity = 0

Acceleration = -4.50 m/s²

Put the value in the equation

Hence, The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

statement #3 on andre"s list is false.