Numerical Problems Based on Instantaneous Velocity and Instantaneous Acceleration for Class 11 Physics

by
  • Last modified on:4 weeks ago
  • Reading Time:8Minutes
Home CBSE Class 11 Physics Numerical Problems Numerical Problems Based on Instantaneous Velocity and Instantaneous Acceleration for Class 11 Physics

Numerical Problems on Kinematics (Calculus Method)

Class 11 Physics · Motion in a Straight Line
SQ

Differentiation in Kinematics Practice Set

Apply the concepts of $v = \frac{dx}{dt}$ and $a = \frac{dv}{dt}$
Ch 3 · Motion in a Straight Line
1
The displacement $x$ of a particle at time $t$ along a straight line is given by $x = \alpha – \beta t + \gamma t^2$. Find the acceleration of the particle.
Answer $2\gamma$ 📝
Detailed Solution

Given the displacement equation:

$$x = \alpha – \beta t + \gamma t^2$$

Velocity ($v$) is the first derivative of displacement with respect to time:

$$v = \frac{dx}{dt} = \frac{d}{dt}(\alpha – \beta t + \gamma t^2)$$
$$v = 0 – \beta(1) + \gamma(2t) = -\beta + 2\gamma t$$

Acceleration ($a$) is the first derivative of velocity with respect to time (or the second derivative of displacement):

$$a = \frac{dv}{dt} = \frac{d}{dt}(-\beta + 2\gamma t)$$
$$a = 0 + 2\gamma(1) = 2\gamma$$
2
The displacement $x$ of a particle is dependent on time $t$ according to the relation: $x = 3 – 5t + 2t^2$. If $t$ is measured in seconds and $x$ in metres, find its (i) velocity at $t = 2\text{ s}$ and (ii) acceleration at $t = 4\text{ s}$. [Central Schools 13]
Answer (i) $3\text{ ms}^{-1}$ (ii) $4\text{ ms}^{-2}$ 📝
Detailed Solution

Given the displacement equation:

$$x = 3 – 5t + 2t^2$$

(i) Velocity at $t = 2\text{ s}$:
Velocity is the derivative of displacement:

$$v = \frac{dx}{dt} = \frac{d}{dt}(3 – 5t + 2t^2) = -5 + 4t$$

Substitute $t = 2\text{ s}$ into the velocity equation:

$$v_{(t=2)} = -5 + 4(2) = -5 + 8 = 3\text{ ms}^{-1}$$

(ii) Acceleration at $t = 4\text{ s}$:
Acceleration is the derivative of velocity:

$$a = \frac{dv}{dt} = \frac{d}{dt}(-5 + 4t) = 4\text{ ms}^{-2}$$

Since the acceleration is constant, it does not depend on time. Thus, at $t = 4\text{ s}$, the acceleration remains $4\text{ ms}^{-2}$.

3
The displacement $x$ of a particle along X-axis is given by $x = 3 + 8t + 7t^2$. Obtain its velocity and acceleration at $t = 2\text{ s}$.
Answer Velocity: $36\text{ ms}^{-1}$, Acceleration: $14\text{ ms}^{-2}$ 📝
Detailed Solution

Given the displacement equation:

$$x = 3 + 8t + 7t^2$$

Velocity ($v$):

$$v = \frac{dx}{dt} = \frac{d}{dt}(3 + 8t + 7t^2) = 8 + 14t$$

At $t = 2\text{ s}$, the velocity is:

$$v_{(t=2)} = 8 + 14(2) = 8 + 28 = 36\text{ ms}^{-1}$$

Acceleration ($a$):

$$a = \frac{dv}{dt} = \frac{d}{dt}(8 + 14t) = 14\text{ ms}^{-2}$$

Since acceleration is constant, at $t = 2\text{ s}$, it remains $14\text{ ms}^{-2}$.

4
The distance traversed by a particle moving along a straight line is given by $x = 180t + 50t^2$ metre. Find: (i) the initial velocity of the particle (ii) the velocity at the end of $4\text{ s}$ and (iii) the acceleration of the particle.
Answer (i) $180\text{ ms}^{-1}$ (ii) $580\text{ ms}^{-1}$ (iii) $100\text{ ms}^{-2}$ 📝
Detailed Solution

Given the position equation:

$$x = 180t + 50t^2$$

Velocity ($v$) is given by differentiating position with respect to time:

$$v = \frac{dx}{dt} = \frac{d}{dt}(180t + 50t^2) = 180 + 100t$$

(i) Initial velocity:
Initial velocity is the velocity at $t = 0$.

$$u = v_{(t=0)} = 180 + 100(0) = 180\text{ ms}^{-1}$$

(ii) Velocity at the end of $4\text{ s}$:
Substitute $t = 4\text{ s}$ into the velocity equation:

$$v_{(t=4)} = 180 + 100(4) = 180 + 400 = 580\text{ ms}^{-1}$$

(iii) Acceleration of the particle:
Acceleration ($a$) is the derivative of velocity with respect to time:

$$a = \frac{dv}{dt} = \frac{d}{dt}(180 + 100t) = 100\text{ ms}^{-2}$$

Related Posts

Also check

Class-wise Contents

Leave a Reply

Join Telegram Channel

Editable Study Materials for Your Institute - CBSE, ICSE, State Boards (Maharashtra & Karnataka), JEE, NEET, FOUNDATION, OLYMPIADS, PPTs

Download Now!

Discover more from Gurukul of Excellence

Subscribe now to keep reading and get access to the full archive.

Continue reading