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

Byju's Answer

Standard IX

Physics

Derivation of Position-Time Relation by Graphical Method

A particle st...

Question

A particle starts to move in a straight line from a point with velocity $10 m s^{- 1}$ and acceleration $- 2.0 m s^{- 2}$ . Find the position of the particle at $t = 5 s$

25 m

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

- 25 m

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

50 m

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

- 50 m

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

Open in App

Solution

The correct option is A $25 m$
Initial Velocity, $u = 10 m s^{- 1}$
Acceleration, $a = - 2 m s^{- 1}$
Time, $t = 5 s$
Displacement, $s$

Using Newton's second equation of motion,
$s = u t + \frac{1}{2} a t^{2}$
$s = (10 \times 5) - (\frac{1}{2} \times 2 \times 5^{2})$
$s = 25 m$

Suggest Corrections

Similar questions

Q. A particle starts to move in a straight line from a point with velocity

10 m s^{- 1}

and acceleration

- 2.0 m s^{- 2}

. Find position(S) and velocity(v) of the particle at

t = 5

Q. A particle start moving along a straight line path with a velocity

10 m s^{- 1}

. After

5 s

, the distance of the particle from the starting point is

50 m

. Which of the following statement about the nature of motion of the particle are correct?

A particle is moving in a straight line with initial velocity of 10 $\frac{m}{s}$ . Acceleration - Time graph of the particle is shown in the figure. Find velocity at t = 10 seconds.

Q. A particle starting from rest moves with constant tangential acceleration on a circular path of radius

\frac{10}{π} m

. After

2.5

rotation the velocity of the particle is

50 m/s

, its tangential acceleration is

Q. Starting from rest a particle moves in a straight line with acceleration

a = (25 - t^{2})^{1 / 2} m / s^{2} f o r 0 \leq t \leq 5 s

a = \frac{3 π}{8} m / s^{2} f o r t \geq 5 s

The velocity of particle at t = 7s is:

Join BYJU'S Learning Program

A particle starts to move in a straight line from a point with velocity 10ms−1 and acceleration −2.0ms−2. Find the position of the particle at t=5s

The correct option is A 25mInitial Velocity, u=10 ms−1Acceleration, a=−2 ms−1Time, t=5 sDisplacement, sUsing Newton's second equation of motion,s=ut+12at2s=(10×5)−(12×2×52)s=25 m

A particle starts to move in a straight line from a point with velocity $10 m s^{- 1}$ and acceleration $- 2.0 m s^{- 2}$ . Find the position of the particle at $t = 5 s$

The correct option is A $25 m$
Initial Velocity, $u = 10 m s^{- 1}$
Acceleration, $a = - 2 m s^{- 1}$
Time, $t = 5 s$
Displacement, $s$

Using Newton's second equation of motion,
$s = u t + \frac{1}{2} a t^{2}$
$s = (10 \times 5) - (\frac{1}{2} \times 2 \times 5^{2})$
$s = 25 m$