Question

the displacement of a body moving along x axis depends on time as √x = t + 1
then the velocity of the body :

(a) Increases with time
(b) Decreases with time
(c) Independent of time
(d) First increases and then decreases


class 11 physics cbse

Answer 1

Thanks for asking the question!

ANSWER::

√x = t + 1

Squaring both sides

(√x)² = (t + 1)²

x = t² + 2t + 1

Now , lets apply differentiation

dx/dt = (d/dt)(t² + 2t + 1) = 2t + 2

Now , we will put some values in t to check which option is right.

We won't take t values negative because time cannot be negative.

v(t) = 2t + 2

v(0) = 2(0) + 2 = 2

v(1) = 2(1) + 2 = 4

v(2) = 2(2) + 2 = 6

After putting values we can see that the velocity is increasing as time increases .

(a) Increases with time will be the answer.

Hope it helps!

Answer 2

S(t) :- √x = t+1

Squaring both sides x = (t+1)² = t²+1+2t

Differentiating both sides w.r.t (t)

dx/dt = 2t+2

Let select any random values.. for example 0,1

Apply :- If x > y & f(x) > f(y)

It implies that function is increasing function.

Similarly if x<y & f(x)<f(y)

It implies that function is decreasing function.

Now put t = 0

We get x = 2

Now put t = 1

We get x = 4

Conclusion :- the function is an increasing function in interval [0,∞)

Hence the body is accelerating or it's velocity will increase with time.

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Put dx/dt = 0

We get t = -1

Hence it's a point of Maxima or minima.

But plotting it wil not make any sense coz we can't take time as negative.