From above graph choose correct option for $t_{1}$ to $t_{2}$ where $t_{1}$ is the time at which velocity of particle = 0 $t_{2}$ is the time at which velocity of particle is negative.

$t_{1} = t_{B}; t_{2} = t_{C}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$t_{1} = 10; t_{2} = 2$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$t_{1} = 10; t_{2} = 17$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$t_{1} = 2; t_{2} = 17$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$t_{1} = 10; t_{2} = 17$

Particle seems to be at rest at t = 10 sec as even before and after t = 10 sec mark the particle continues to be at 10m displacement from starting point. Thus its displacement doesn't change.

So at t = 10 sec, velocity = 0

Now for the velocity to be negative

Let's analyze the motion of the particle: till time particle moved away from starting point and reached a distance of 10m.

Till $t = t_{c}$ particle stayed there maintaining it displacement of 10m.

Now after $t_{c}$ the particle starts moving towards the starting point as its displacement is decreasing.

So particles velocity was leftwards after $t = t_{c}$

$\Rightarrow$ Particle was moving with a negative velocity in the time instance t = 17 sec, 20 sec, 30 sec etc.

Suggest Corrections

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.

From above graph choose correct option for $t_{1}$ to $t_{2}$ where $t_{1}$ is the time at which velocity of particle = 0 $t_{2}$ is the time at which velocity of particle is negative.

Q. A particle is projected with speed

u

at angle

θ

with horizontal from ground. If it is at same height from ground at time

t_{1}

and

t_{2}

, then its average velocity in time interval

t_{1}

t_{2}

Join BYJU'S Learning Program

From above graph choose correct option for t1 to t2 where t1 is the time at which velocity of particle = 0 t2 is the time at which velocity of particle is negative.

From above graph choose correct option for $t_{1}$ to $t_{2}$ where $t_{1}$ is the time at which velocity of particle = 0 $t_{2}$ is the time at which velocity of particle is negative.