1) A truck travelling due north at 50 km/h turns west and travels at the same speed.What is the change in ....

1♠ vector change = final – initial:

Δv= -50*i –50*j = 50*sqrt2(-i*cos45° -j*sin45°); C;

2♣ A; according to definition of cross product it may be parallel or anti-parallel to F1;

3♥ v1= 100*j; s1=j*200km; v2=100*i; s2=i*200km; the graph for acceleration looks:

a=dv1/dt=0 for 0 < t < 2 hours;

a=dv2/dt=0 for 2h < t < 4h;

yet for t=2h, a=(v2-v1)/ Δt is indefinite as Δt → 0; D

This is a multiple question, so I will answer in multiple parts:

1)C The question is that of resultant change. If your change is from 50km/h north to 0 km/h north, your change in velocity would be 50km/h south. If however you are changing from 0km/h west to 50km/h west, the change would be 50 km/h west. Now the resultant is direction southwest with triangular calculation 50 square plus 50 square and take the square root thereof, which is 5000 squareroot. or 50 squareroot 2 if you whish.

2) If a vector product is zero, then the two vectors must me perpundicular to one another. Now in this question, I would ask you, what are the axis systems. Find the axis that is perpendicular to your axis and that would be your answer.

3) B. Similar question and method of question 1. Ignore the distance travelled as it has nothing to do with the aceleration of the body. Then take the 100's squared ,add them and take the square root thereof.

Good luck.

1) v(resultant) = v(final) - v(initial)

= (0i-50j) - (50i+0j)

= -50i-50j

or 50sqrt(2) north - west (not in the options, but this is the answer...)

2) f1 = +i

f1.f2 = f1xf2xsin(theta) = 0 (=>theta=0)

therefore f2 is parallel to f1 =>f2 = 4i

3) the answer is D)none of these, because, the change in velocity(by changing directions, a change in velocity was produced, as in Q1) was instantaneous, so assuming t=0hr.

acc.(a)=(change in velocity)/time(t)

=>a=100sqrt(2)/0

=>a=infinitykm/h^2

substitute in velocity is named acceleration. whilst the truck is shifting north with uniform velocity, there is not any substitute in velocity. there is an application of tension that alterations its direction west wards. in this era the northward velocity or velocity is delivered to 0 and the automobile is speeded as much as head west wards till the cost reaches 30m/s. Afterwards there is not any velocity substitute.

100 km/h^2

i dont know.