i need help with an ap physics 1 frq?
here is the frq, the attached image is the graph it refers to:

two identical objects, a and b, move along straight, parallel, horizontal tracks. the graph attached as the image represents the position as a function of time for the two objects.

(a) at approximately which time or times, if any, are the objects moving at the same speed? if the objects are never moving with the same speed during the interval shown, indicate this explicitly. briefly explain your answer.

(b) describe what is happening to the positions of objects a and b between 3 seconds and 4 seconds.

(c) which object, if either, has an acceleration with a greater magnitude during the time interval shown in the graph? if the accelerations have the same magnitude for both objects, indicate this explicitly.

(d) in a clear, coherent, paragraph-length response, explain your response to part (c). be sure to reference and compare the graph information for both objects a and b.
(a) to my eye, they never have a common speed. that would require the two curves to have the same slope at the same time, which doesn't appear to me to be true.

(b) b is ahead of a but is slowing down and being passed by a, which has been moving at constant speed.

(c) b has an acceleration with greater magnitude. (in fact, the curve for a doesn't show any acceleration at all.)

(d) a has a constant slope, and therefore a constant velocity (and therefore displays no acceleration at all). b starts ahead of a initially with a velocity greater, but then decelerates as it nears t = 4 s. at around t = 3.7 s, b's velocity (slope) has nearly dropped to zero, and it gets passed by a.

if you find this helpful, please select favorite answer!
hint: here, we will proceed by finding out the relative speed between the two cars and then using the formula i.e., time taken$ = \dfrac{{{\text{distance travelled}}}}{{{\text{speed}}}}$ to determine after how many hours these cars will meet each other.complete step-by-step answer:given, distance between the two cars d=160 kmstarting time for both the cars when they left their respective initial positions a and b is 8:10 amspeed of the first car, x=50 km per hourspeed of the second car, y=30 km per hourin order to obtain the relative speed between these two cars which are moving towards each other (i.e., in opposite directions), the individual speeds of these cars will be added.so, the relative speed between the two given cars, r= x+y=50+30=80 km per houras we know that, time taken$ = \dfrac{{{\text{distance travelled}}}}{{{\text{speed}}}}$so, time when both the cars will meet=$\dfrac{{{\text{initial distance between both the cars}}}}{{{\text{relative speed between the cars}}}} = \dfrac{{\text{d}}}{{\text{r}}} = \dfrac{{160}}{{80}} = 2$ hours.after 8:10 am, both the cars which are initially at different places a and b will meet each other in the next 2 hours.therefore, time when they will meet each = 8:10 am + 2 hours = 10:10 amhence, option a is correct.note: in this particular problem, since the given two cars are moving in opposite directions (or moving towards each other) that’s why the individual speed of the cars are added in order to determine the relative speed between them but if they were moving in the same direction then the relative speed between these cars would have been obtained by subtracting the individual speed of these cars. physics 101 notes: 8/25 (week 2 continued)  force o defined as a push or a pull, and is a vector quantity, meaning that it has direction depending on which way it is being pushed or pulled o it has amount and direction (how much force is being applied and in which direction it is moving)  newton’s first law o first draft: “an object that is free of forces move in a straight line and covers equal distances in equal time” o final: “an object that is free of external forces moves at a constant velocity”  how does a skater start, stop, or turn? o acceleration: “the rate at which an object’s velocity is changing with time o acceleration, s (a) g(x) is a horizontal translation of 2 units to the left then a reflection over the x-axis of f(x)
  

(b) h(x) is a horizontal translation of 3 units to the right then a vertical translation os 4 units up of graph of f(x). ?solution 
step 1:
we have to draw the graph of form r to r. 
step 2:
the graph of f is the set of all points (x,f(x)) with x in the domain of f.
we take x in [-4,4]
and we will find the separate values of y for separate intervals.
 
 
 

 
......(1)
 

 

 

using x values along x-axis and along y-axis we get the graph of given function form r to r.
step 3:
the graph is as follow yes. let us consider a block a which is resting on another block b. block b is resting on a smooth horizontal surface. let the coefficient of friction between the blocks be \[\mu\] . when a force f is applied on block b in the forward direction as shown in the above figure, block a moves with block b in the direction of the applied force. the frictional force on block a and the displacement will be in the forward direction. therefore, work done by the frictional force is positive. if we consider the reference frame of block b, then displacement of block a will be zero. therefore, work done by the frictional force is zero. the graph of this equation is a line.  we know that because each term is either a constant (1.5) or the product of a constant and the first power of a variable (d, 0.8t).
since we have two variables (d and t), the line sits in a two-dimensional plane, with a vertical axis and a horizontal axis.  d stands alone in the equation, so its value depends on the value oft.  therefore, the vertical--or dependent--axis will represent values of d, and the horizontal--or independent--axis will represent values of t.
at this point, then, we can set up a coordinate plane with a vertical axis labeled d and a horizontal axis labeled t.  all of the points on our plane have a coordinate (t, d).
to graph a line, we need two points.  since the question asks for the axial intercepts anyway, we can use those.
we'll start with the d-intercept, where the line crosses the d-axis.  all of the points on the d-axis have a t-coordinate of zero, so to find where the line crosses the axis, we substitute zero for t in our equation and solve.
d = 1.5 + 0.8 × t
d = 1.5 + 0.8 × (0)
d = 1.5 + 0
d = 1.5
so our d-intercept has the coordinates (0, 1.5), and we have our first point and our first intercept.
similarly, to find the t-intercept, we note that all of the points on the t-axis have a d-coordinate of zero, so we set d equal to zero and solve.
d = 1.5 + 0.8 × t
0 = 1.5 + 0.8 × t
0 - 1.5 = 1.5 - 1.5 + 0.8 × t
-1.5 = 0.8 × t
-1.5 ÷ 0.8 = (0.8 ÷ 0.8) × t
-1.875 = 1 × t
-1.875 = t
and the d-intercept is (-1.875, 0).
now we have both axial intercepts, so that part of the question is complete.  we can plot those two points and draw the line that passing through them (extending on, at either side, to infinity).
all that remains is to find the turning points.  a turning point, also called a local maximum or local minimum, is where the value of a function goes from decreasing to increasing or from increasing to decreasing.  (the graph shown at http://en.wikipedia.org/wiki/file:parabola\_with\_focus\_and\_directrix.svg has a minimum at the origin, for example.)
a line has a constant slope.  to say that differently, the angle between any two points on a line is the same as the angle between any two other points on the line.  if a line is increasing in one location (like our line), it is increasing everywhere, and it is increasing everywhere at the same rate.  therefore, a line has no turning points. u (initial velocity) = 18 km/hr 
t ( time )= 2.5 sec
v( final velocity) = 0 m/s
from, the first equation of motionv = u +  a×t0 = 5 +  a×2.5a = - 2 m/s2 Physical fitness is the total functional capacity of an individual to perform a given task. It means that it is the body’s ability to function efficiently, to enjoy leisure time, to be healthy, to resist hypokinetic diseases and to meet unforeseen situation. It consists of health related fitness and skill related physical fitness. Physical fitness and exercises habits developed in early years provide a foundation for life time. Since activity is the basis of life and human body cannot remain in normal condition without activity therefore optimal physical fitness is not possible without regular exercise.

**Components of physical fitness**  
**There are five physical fitness components. These are  
- Speed  
- Strength  
- Endurance  
- Flexibility  
- Coordination**  
- Speed : It is the ability to perform movement at faster rate or. It is the ability to perform movement in a short period of time and sports  
e.g. practicing with faster rhythm, speed endurance repetition method, acceleration runs etc.  
- Strength : It is an ability of muscles to overcome or to act against resistance e.g. exercise with medicine ball, Isometric exercises, pushups etc.  
- Endurance : It is the ability to sustain or continue activity or it the ability to resist fatigue. It is one of the important components for middle and long distance races ad it is required for almost all major games like football, hockey and basketball.

- Flexibility : It is the capacity of a muscle to extend without any damage. Flexibility is measured by range of motion around a joint. It is affected by muscle length, joint structure and other factors. It is measured through flexometer.  
- Co-ordinate ability : It is the ability of the body to perform movement with perfection and efficiency. In other words it is ability to change movement or direction in the shortest time without getting unbalanced. &nbsp;an electron enters uniform electric field maintained by parallel plates and of value 'e' &nbsp; v m-1 with a velocity ' v ' ms-1.&nbsp;the plates are separated by a distance 'd' metre. what is the acceleration of the electron in the field?&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; answer: refractive index is defined as the ratio of the speed of light in a vacuum to its speed in a specific medium. refractive index is dimensionless. it is a number that indicates the number of times slower than a light wave would be in the material than it is in a vacuum. refractive index, represented by symbol n, is the velocity of light in vacuum divided by the velocity of light in a medium. the formula of the refractive index is, n = c/ν where n = refractive index. c = velocity of light in a vacuum ( 3 × 108 m/s). v = velocity of light in a substance. snell’s law clarifies the relation between the angle of incidence and angle of refraction. so, there are two formulas for calculating the refractive index of a medium. let us imagine that a ray of light is travelling from a medium a to another medium b. then conferring from snell’s law, .