PHYSICS [3 Mark Question Answer]

(i) The body is showing decreasing velocity that is retardation.
(ii) Initially, the body is showing retarded motion and then an accelerating one.
(iii) The body is in the state of rest.

Body is travelling with uniform velocity from point A to B i.e for (40 - 20) = 20 hr.

{a) Speed as it moves from o to a = {6 - O)/ {3 - 0 = 2 ms-1.
(b) Speed as it moves from A to B = {6 - 6)/ (5 - 3) = o ms-1.
(c) Speed as it moves from B to c = {14 - 6)/ {8 -5) = 8/ 3 = 2.66 ms-1

8 travel 4 km between the time he passed C and A.

C is at 8 km when B meets A.

All of them never come at the same point at the same time.

8 is travelling fastest among all the 3 students

Car travls 30 km distance with speed 60 km/hr
Time taken by car to travel this distance = 30/60 = 0.5 hr.
Car travels another distance of 30 km with speed of 20 km/hr.
Time taken by car to travel this distance = 30/20 = 1.5 hr.
Total time taken = 1.5 + 0.5 = 2 hr.
Total distance = 30+ 30 = 60 km.
Average speed of car = 60/2 = 30 km/hr.

Angular displacement θ = length of arc/ radius of circle
θ= I / r
I= θxr
divide above equation with t
l/t = θxr/ t
now l/t = v {linear speed)
θ/ r= w
So v = ωr.

As acceleration is the slope of line in velocity time graph and as B has greater slope than line A.
So B has greater acceleration than A.

Circumference of track = 314 m.
We know that circumference = 2nr
So 2nr = 314
R = 314/ 2n = 50 m
Diameter of track would be = 2 x 50 = 100 m.
Length of path AB= 100 m.
(a) Distance moved by cyclist= half the circumference of track i.e 314/ 2 = 157 m.
(b) The displacement of cyclist is equal to the diameter of circle i.e AB= 100 m.
(c) Average velocity would be equal to = 15.7 ms-1.

Radius of orbit is 42250 km.
Distance covered by satellite to complete 1 revolution is 2 πr.
So distance is = 2 X 3.14 X42250 =265330 km.
time taken by satellite to complete one revolution is 24 hr = 24X60X60=86400 sec.
so linear velocity is= distance /time= 265330/86400 ms-1 = 3.07 km s-1.

ln this graph portion O to A represents motion with acceleration, A to E represents motion with uniform velocity, 8 to C represents motion with acceleration.

In distance time graph a straight line parallel to time axis represents state of motion.

As slope of velocity time graph gives acceleration. So motion with zero acceleration is represented by line having zero zero slope with time axis or a line parallel to the time axis.

Deceleration is represented by a straight line having a negative slope with time axis.

Acceleration is represented by a line on a velocity time graph with a positive slope with time axis.

As slope of this line is negative so this represent decreasing velocity.

As uniform velocity means slope of line should be constant throughout the motion so this graph also represents uniform velocity

As slope of a distance time graph indicates velocity so a increasing velocity means a straight line having a positive slope with time axis.

As slope of line A is greater than line 8 that means velocity of body A is greater than body 8 or in other words body A is moving faster

Graph (c) represents a motion with negative acceleration.

If displacement time graph of a body is straight line moving upwards and starting from origin then it means body has started from and moving with a uniform velocity,

If distance time graph of a body is straight line parallel to x axis then the body is said to be at rest .

Spped time graph for a uniformity retarded motion

Speed time graph for a body showing fluctuating speed.

Distance time graph for a body at rest.

Speed time graph of a body starting from point P gradually picking up speed, then running at a uniform speed and finally slowing down to stop at some point Q

Speed time graph of a body when itS Initial speed is not ze ro and and speed increase uniformly with time.

Slope of a graph is given as rate of change of y coordinates to the x coordinate. In speed time graph speed is on the y axis and time is on the x axis. And we define acceleration as rate of change of speed with respect to time. So slope of a speed time graph gives acceleration.

Yes the motion is uniform and the uniform speed is given by area under speed time graph divided by time interval.
So speed = 500/20 =25 m/s.

If the acceleration of a moving body is constant in magnitude and direction then the path of the body must not be a straight line because in circular motion also acceleration of a body is constant in magnitude and always directed towards the centre.
So the path of the body may be a straight line and may be a circular one.

We convert all the speeds in m/s to compare them.
36 km/hr = (36 X 1000)/3600 = 10m/s.
2 km/min = (2 X 1000)/60 = 33.3 m/s.
7 m/s = 7 m/s.
So increasing order of speed is 7m/s < 10m/s <33m/s.