Could you show me how to do these physics problems?

1. A car with a velocity of 25 m/s comes to rest in 10 seconds by applying brakes with a constant braking force. How far does the car travel in those 10 seconds?

S = 1/2*Vi*t = 25*5 = 125 m

You dont need to know the braking force !!!! available data are more than enough !!!!

2. A baseball is thrown at an angle of 30° with respect to the ground and it reaches the ground in 2 seconds. What is its initial velocity of baseball?

The answer depends upon the height the ball is thrown from ....nothing is said on the matter...so a couple of calculation are carried out to show the procedure

a) h = 0

h = 0 = Vsin30*2-4.9*2^2

4.9*4 = V = 19.6 m

b) h = 1

h = -1 = Vsin30*2-4.9*2^2

19,6-1 = V = 18.6 m/sec

The sign minus relevant to the height h shows that landing height is below the throwing point

3. A car accelerates uniformly from rest on a straight road to a speed of 30 m/s in 15 seconds. What is the distance covered in 10 seconds?

S = 1/2Vf*10/15*t = 10*10 = 100 m

1 The car slows from 25m/s to 0 in 10s.

The fact that it is caused by a 'braking force' is irrelevant.

Average velocity = (25+0)/2 = 12.5m/s

Distance = average velocity x time = 12.5 x 10 = 125m

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2 You are correct. I would have done it a different way though.

Find the initial velocity of an object thrown vertically, if it takes 2s to come back down.

vf = vi + at

vf = -vi

-vi = vi + (-9.8)(2)

2vi = 19.6

vi = 9.8m/s

The ball's vertical component of velocity must also be 9.8m/s as it also takes 2s to rise and fall.

If the initial velocity is V, then :

Initial vertical component of velocity = Vsin(30⁰)

Vsin(30⁰) = 9.8

V = 9.8/(sin(30⁰) = 19.6m/s

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3 You can't multiply acceleration by time and get distance.

So you made a mistake when you wrote: 2m/s^2 (10) = 200m

a = (vf - vi)/t = (30-0)/15 = 2m/s² (agreed)

d = vi.t + ½at ² (a standard formula you need)

= 0x10 + ½ x 2 x 10²

= 100m