You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Physics

Distance-Time Graphs

The distance ...

Question

The distance - time graph of a particle at time t makes an angle 45 $^{c} i r c$ with time axis. After 1 second, it makes angle 60 $^{c} i r c$ with time axis. What is the acceleration of the particle?

$\sqrt{3} - 1$ unit

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

$\sqrt{3} + 1$ unit

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

$\sqrt{3} + 1$ unit

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

$1$ unit

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

Open in App

Solution

The correct option is A
$\sqrt{3} - 1$ unit

The slope of the distance - time graph gives speed.
The change in velocity in 1 second.
$= t a n 60^{\circ} - t a n 45^{\circ} = \sqrt{3} - 1$
So, acceleration
$\Rightarrow \frac{Δ v}{Δ t} = \frac{\sqrt{3} - 1}{1} = (\sqrt{3} - 1)$

Suggest Corrections

Similar questions

Q. The distance (

m

) vs. time (

t

) graph of a particle at

t = 0

makes an angle of

45^{0}

with time axis. After

1

second it makes an angle of

60^{0}

with time axis, what is the acceleration (in

m / s^{2}

) of the particle, assuming its acceleration is constant for given time period?

Q. If a particle starts moving along x-axis from the origin with initial velocity

u = 1 m s^{- 1}

and acceleration

a = 2 m s^{- 3}

, the relationship between displacement and time is

x = u t + \frac{1}{2} a t^{2} = 1 \times t + \frac{1}{2} \times 2 \times t^{2} = t + t^{2}

Draw the displacement

(x) - t i m e (t)

graph.

Q. Tangential acceleration of a particle moving in a circle of radius

1 m

varies with time

t

as shown in the figure (initial velocity of particle is zero). Time after which total acceleration of particle makes an angle of

30^{\circ}

with radial acceleration is

Q. Displacement - time curve of a particle moving along a straight line is shown. Tangents at

A

and

B

make angles

45^{\circ}

and

135^{\circ}

with positive x-axis respectively. The average acceleration of the particle during

t = 1, t = 2

second is

Q. Tangential acceleration of a particle moving in a circle of radius

1 m

varies with time

t

as (initial velocity of particle is zero). Time after which total acceleration of particle makes and angle of

30^{o}

with radial acceleration is

Join BYJU'S Learning Program

The distance - time graph of a particle at time t makes an angle 45circ with time axis. After 1 second, it makes angle 60circ with time axis. What is the acceleration of the particle?

The correct option is A √3−1 unit The slope of the distance - time graph gives speed. The change in velocity in 1 second. =tan60∘−tan45∘=√3−1 So, acceleration ⇒ΔvΔt=√3−11=(√3−1)

The distance - time graph of a particle at time t makes an angle 45 $^{c} i r c$ with time axis. After 1 second, it makes angle 60 $^{c} i r c$ with time axis. What is the acceleration of the particle?

The correct option is A
$\sqrt{3} - 1$ unit

The slope of the distance - time graph gives speed.
The change in velocity in 1 second.
$= t a n 60^{\circ} - t a n 45^{\circ} = \sqrt{3} - 1$
So, acceleration
$\Rightarrow \frac{Δ v}{Δ t} = \frac{\sqrt{3} - 1}{1} = (\sqrt{3} - 1)$