Question

plzz rєαd thє quєѕtíσnѕ cαrєfullч αnd αnѕwєr.....


ⓐⓝⓢⓦⓔⓡ ⓐⓒⓒⓞⓡⓓⓘⓝⓖ ⓣⓞ ⓜⓐⓡⓚⓢ!!!!!!!

αnѕwєr thє quєѕtíσn чσu knσw...



plєαѕє í urgєntlч nєєd thє hєlp



ít íѕ vєrч ímpσrtαnt!​

Answer 1

Q7. (a)When an object is thrown upwards it's velocity decreases continuously till zero then finally the object drops(freefall) down with increasing velocity.....

In this process, time continuously increases... So, the 1st graph is according to the condition given.

(b)An object is decelerating that means velocity is decreasing... it attains a velocity lower than before.... then it starts accelerating again.During this, the time passes by... So, last graph suits the condition well...

Yes i must ads on that 1st graph bhi ho sakta hai

but 1st graph already 1st condition k liye hai so, last graph hi answer hoga.

(c)Uniform motion means no force (accelerating or decelerating) hence velocith remains constant with time....

So, second graph suits te condition.

Answer 2

8] When the speed or velocity of an object moving along a straight line remains constant, its motion is said to be uniform motion. Thus, in uniform motion, the speed/velocity of a body is constant and the direction of motion does not change. Clearly, we can say when a body covers equal distances in equal intervals of time, then the motion is uniform motion.

For example, a train running in specific direction at a constant speed/velocity, a car moving along a straight road at constant speed, etc.

The reverse case happens in case of non-uniform motion. In non - uniform motion, the velocity (or speed) of a body changes with time. Clearly, we can say when a body covers equal distance in unequal intervals of time or unequal distance in equal intervals of time, then the motion is non-uniform.

For example, a body moving in a circular path, a bus moving on a busy road, etc.

9] Acceleration is the rate of change of velocity of a body. When the change in velocity by a body done with respect to a certain time, we term it as acceleration.

We must know the formula that :

$a = \frac{v - u}{t}$

Here, "a" is acceleration produced in a body, "v" is the final velocity acquired by a body, "u" is the inital velocity with which the body was actually moving and "t" is the total time taken.

(A) Let us consider a body starting from rest. That means its initial velocity was 0 m/s. Now, it is accelerating and accelerating and its velocity is actually increasing from 0 m/s. Let it changes its velocity from 0 m/s to 10 m/s at time 5s. So, its acceleration will be :

$a = \frac{v - u}{t} = \frac{10 - 0}{5} = \frac{10}{5}$

Hence, a = 2 m/s² (which is positive)

(B) Let us consider a body moving with a certain velocity (say, 4 m/s), which strikes a wooden block certainly and as a result, comes to rest, i.e., final velocity becomes 0 m/s. Let, the total time taken be 2 s. So, its acceleration will be :

$a = \frac{v - u}{t} = \frac{0 -4 }{2} = \frac{ -4}{2}$

Hence, a = -2 m/s² (which is negative)

(C) Consider a body moving with constant speed (say, 4 m/s) and its velocity is not changing. The actual fact is that, no external force is applied on that body to make it accelerate more, i.e., to increase its velocity. Let the total time taken be 2 m/s

So, in this case, no acceleration will be produced in the body and hence, acceleration will be zero.

$a = \frac{v - u}{t} = \frac{4-4 }{2} = \frac{0}{2}$

Hence, acceleration produced on the body is 0 m/s²

Now, practically the examples can also be mentioned as under :

(A) A car initially at rest, increasing its velocity after a certain period of time.

(B) A car moving with certain velocity, suddenly stopped (i.e., comes to rest) after a certain period of time.

(C) A car moving with constant velocity.

10] Initial difference in height = (150 – 100) m = 50 m

Distance travelled by first body in 2 s = h1 = 0 + 1/ 2 g (2)²= 2 g

Distance travelled by another body in 2 s = h2 = 0 + 1/ 2 g (2)² = 2 g

After 2 s, height at which the first body will be = h1 ′ = 150 – 2 g

After 2 s, height at which the second body will be = h2′ = 100 – 2 g

Hence, after 2 s, difference in height = 150 – 2 g – (100 – 2 g) = 50 m = initial difference in height

So, we can say that the difference in height does not vary with time.