I need help in these questions in details

17) is c.

I think all the others have been answered.

the answer
length = + 5
time = +2
v= 5/2 = + 2.5

the length still bositive any way the direction is nigative or not 

Question 5 is right.  

Question 11   There may be a 'normal' method for doing this, I am not a physics specialist.

$$magnitude\\ =\sqrt{7.6^2+5.2^2}\\ =9.20869\\ =9.2$$

$$\\tan\theta=\frac{5.2}{7.6}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{5.2}}}{{\mathtt{7.6}}}}\right)} = {\mathtt{34.380\: \!344\: \!723\: \!845^{\circ}}}$$

direction = 180+34.38 = 214.38 degrees

So the answer is B      9.2 and 214.38 degrees with the positive x axis.

If you do not understand what I have done then make sure you ask me to explain more.  

Thank you aliammar 

Melody .. its very clear .. the mistake that I do un my solution that I took this angle 

Good,

Lets try question 16

easterly force = 3N

Northerly force =4N

Use Pythagoras's theorem to get net force.

sqrt(9+16)=5N

Now,  F=ma

5=2*a

$$acceleration =5/2=2.5m/s^2$$

Question 19

NOTE:    I do not know this stuff.  I am referencing this site

http://zonalandeducation.com/mstm/physics/mechanics/forces/newton/mightyFEqMA/mightyFEqMA.html

It looks pretty good.  

u=30m/s,  mass=2000kg,  Braking force=10000N,  v=0,   find t

F=ma    therefore a=F/m=-10000/2000=-5m/s²

v=u+at

0=30-5t

5t=30

t=6 seconds

------------

d=ut+(1/2)at²

d=30*6+(1/2)(-5)*36

d=180+-90

distance=90metres      I think that D is the answer.

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for 16 .. Now its very clear  thank you 

I don't know question 17 sorry.

for 19 .... also now its very clear 

For question 5) the average velocity is (x(4secs) - x(2secs))/(4secs - 2secs)

This is (5 - 10)/(4 - 2) = -5/2 = -2.5 m/s

So b is the correct answer.

Thank you alan ... waiting for remaining questions 

Thanks Alan.  

Thanks for all ... alan and melody