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Standard XII

Physics

SHM expression

Figure shows ...

Question

Figure shows the displacement-time graph of a particle moving in a straight line. Find the signs of velocity and acceleration of particle at time t= $t_{1}$ and t= $t_{2}$

v_{t_{1}}

and

a_{t_{2}}

are positive while

v_{t_{2}}

is negative

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v_{t_{1}}

and

a_{t_{2}}

are negative while

v_{t_{2}}

is positive

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v_{t_{1}}

and

v_{t_{2}}

are positive while

a_{t_{2}}

is negative

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v_{t_{2}}

and

a_{t_{2}}

are positive while

v_{t_{1}}

is negative

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Solution

The correct option is A $v_{t_{1}}$ and $a_{t_{2}}$ are positive while $v_{t_{2}}$ is negative
slope at $t_{1}$ is positive and that of at $t_{2}$ s negative. Slope at $t_{1}$ is increasing and at $t_{2}$ is decreasing, thus, the acceleration is positive at $t_{1}$ and negative at $t_{2}$

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Similar questions

Q. The velocity-time graph of a particle in one-dimensional motion is shown in Which of the following formulae are correct for describing the motion of the particle over the time-interval t (a) x(t₂ ) = x(t₁) + v (t1) (t₂ – t₁) +(½) a (t₂ – t₁)₂ (b) v(t₂ ) = v(t₁) + a (t₂ – t₁) (c) vaverage = (x(t₂) – x(t1))/(t₂ – t₁) (d) aaverage = (v(t₂) – v(t1))/(t₂ – t₁) (e) x(t₂ ) = x(t₁) + vaverage (t₂ – t₁) + (½) aaverage (t₂ – t₁)₂ (f) x(t₂) – x(t₁) = area under the v-t curve bounded by the t-axis and the dotted line shown.

The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:

Which of the following formulae are correct for describing the motion of the particle over the time-interval t₂ to t₁?

(a) x(t₂) = x(t₁) + v(t₁)(t₂–t₁) + ()a(t₂–t₁)²

(b) v(t₂)= v(t₁) + a(t₂–t₁)

(d) a_Average = (v(t₂) – v(t₁)) / (t₂ – t₁)

(e) x(t₂) = x(t₁) + v_Average(t₂ – t₁) + ()a_Average(t₂ – t₁)²

(f) x(t₂) – x(t₁) = area under the v–t curve bounded by the t-axis and the dotted line shown.

Q. For any arbitrary motion in space, which of the following relations are true:
(a)

v_{a v e r a g e} = (1 / 2) (v (t_{1}) + v (t_{2}))

(b)

v_{a v e r a g e} = [r (t_{2}) - r (t_{1})] / (t_{2} - t_{1})

(c)

v (t) = v (0) + a t

(d)

r (t) = r (0) + v (0) t + (1 / 2) a

t^{2}

(e)

a_{a v e r a g e} = [v (t_{2}) - v (t_{1})] / (t_{2} - t_{1})

(The 'average' stands for average of the quantity over the time interval

t_{1}

t_{2}

)

Q. For any arbitrary motion in space, which of the following relations are true : (a) v _average = (1/2) (v (t₁) + v (t₂)) (b) v _average = [r(t₂) - r(t₁) ] /(t₂ – t₁) (c) v (t) = v (0) + a t (d) r (t) = r (0) + v (0) t + (1/2) a t₂ (e) a _average =[ v (t2) - v (t₁ )] /( t₂ – t₁) (The ‘average’ stands for average of the quantity over the time interval t₁to t2)

Q. Read the following statements and state whether they are True or False:
i) If

u

and

a

both are negative, motion is only retarded.
ii) If

u

is negative but

a

is positive, then displacement of the particle can never be positive.
iii) If u is positive but

a

is negative particle comes to rest for a moment at some time but if

u

is negative and

a

positive it never comes to rest.

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Figure shows the displacement-time graph of a particle moving in a straight line. Find the signs of velocity and acceleration of particle at time t=t1 and t=t2

The correct option is A vt1 and at2are positive while vt2is negativeslope at t1 is positive and that of at t2 s negative. Slope at t1 is increasing and at t2 is decreasing, thus, the acceleration is positive at t1 and negative at t2

Figure shows the displacement-time graph of a particle moving in a straight line. Find the signs of velocity and acceleration of particle at time t= $t_{1}$ and t= $t_{2}$

The correct option is A $v_{t_{1}}$ and $a_{t_{2}}$ are positive while $v_{t_{2}}$ is negative
slope at $t_{1}$ is positive and that of at $t_{2}$ s negative. Slope at $t_{1}$ is increasing and at $t_{2}$ is decreasing, thus, the acceleration is positive at $t_{1}$ and negative at $t_{2}$