Motion in a Straight Line MCQ Quiz - Objective Question with Answer for Motion in a Straight Line - Download Free PDF

Explanation:

Given the equation of motion:

x = 4t²

The velocity (v) is the derivative of displacement with respect to time:

v = dx/dt

Taking the derivative of x = 4t²:

v = 8t

At t = 5 sec:

v = 8 × 5 = 40 m/s

Final Answer: v = 40 m/s

Calculation:

Given:

Initial velocity, u = 12 m/s

Final velocity, v = 0 m/s (since the car comes to rest)

Distance, s = 45 m

Using the formula:

v² = u² + 2as

⇒ (0)² = (12)² + 2 × a × 45

⇒ 0 = 144 + 90a

⇒ 90a = -144

⇒ a = -1.6 m/s²

Calculation:

let x distance covered with v₁ speed and other x distance with v₂ speed

Average velocity = Total displacement / Total time

= (x + x) / (x / v₁ + x / v₂)

= (2 × v₁ × v₂) / (v₁ + v₂)

Calculation:

The average velocity is given by:

v_avg = (x₁ + x₂) / (t₁ + t₂)

Given: v_avg = 50 / 7

Let x be the initial distance:

x₁ = x,

x₂ = (3x / 2),

t₁ = x / 5,

and t₂ = (3x / 2) / v₂

⇒ 50 / 7 = (x + 3x / 2) / (x / 5 + (3x / 2v₂))

⇒ 50 / 7 = (5 / 2) / (1 / 5 + 3 / (2v₂))

⇒3 / (2v₂) = 7 / 20 - 1 / 5 = (7 - 4) / 20 = 3 / 20

⇒3 / (2v₂) = 3 / 20

⇒  v₂ = 10 m/s

Solution:

The positions of the particle at t = 0 and t = 1 are:

x(0) = 0 m,    x(1) = 1 m

The particle is at rest at t = 0 and t = 1. Thus, initial and final velocities are:

v(0) = 0,    v(1) = 0      (1)

The velocity v(1) of a particle having instantaneous acceleration α is given by:

v(1) = v(0) + ∫ α dt from 0 to 1      (2)

From equations (1) and (2): ∫₀¹ α dt = 0

If α is always > 0 in the interval, then the area under α-t curve would be positive, contradicting this result. Similarly, if α is always < 0, area would be negative. So, α must change sign in the interval (0, 1).

Hence, only conclusions that can be made are: ∫₀¹ α dt = 0, α ≠ 0, and α must change sign (at least once) in (0, 1).

If α changes sign only once (say at t = tc) and α = α₁ for 0 < t < tc and α = -α₂ in tc < t < 1, where α₁ and α₂ are positive constants, then to maintain total area under v-t graph as 1 m, we must have:

Area = (1/2) × 1 × vmax = 1 ⇒ vmax = 2 m/s

Also, vmax = α₁ × tc = α₂ × (1 - tc)

If tc < 1/2, then α₁ > 2; otherwise α₂ > 4

Hence, at some point |α| must be ≥ 4

Answer: (A), (C)

CONCEPT:

• Equation of motion: The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.
• These equations are only valid when the acceleration of the body is constant and they move on a straight line.

There are three equations of motion:

V = u + at

V2 = u2 + 2 a S

$\text{S}=\text{ut}+\frac{1}{2}\text{a}{\text{t}}^2$

Where, V = final velocity, u = initial velocity, s = distance traveled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

EXPLANATION:

Given that:

Initial velocity (u) = 0

Distance (S) = 20 m

Time (t) = 4 sec

Use $\text{S}=\text{ut}+\frac{1}{2}\text{a}{\text{t}}^2$

20 = 0 + $\frac{1}{2}\times a\times 4^2$

acceleration = a = 20/8 = 2.5 m/s²

The correct answer is (v-u)/t.

CONCEPT:

• Acceleration: The rate of change in velocity is called acceleration. It is denoted by a.
  • The SI unit of acceleration is m/s².
• Equation of motion: The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.
• V = u + at
• Where, V = final velocity, u = initial velocity, a = acceleration of body under motion, and t = time taken by the body under motion.

EXPLANATION:

From the above equation of motion:

V = u + a t

So a = (V - u)/t

• The formula for finding acceleration is (v-u) / t. So option 4 is correct.

Additional Information

• The product of velocity and time (v x t) represents the displacement of a body which represents the shortest distance between two points.
• The reciprocal of time (1 / t) represents the frequency of oscillations which gives the value of the total number of oscillations in the given time.
• The work per unit time (W / t) represents power which is the measurement of the rate of doing work.

Concept:

• Displacement: The change in position is called displacement. It is the shortest distance that can be traveled between two points.
• Velocity: The rate of displacement is called velocity. It is a vector quantity. like displacement.

$\overrightarrow{v}=\frac{\overrightarrow{ds}}{dt}$

s is displacement, t is time.

• Acceleration: The rate of change of velocity is called acceleration.

$\overrightarrow{a}=\frac{\overrightarrow{dv}}{dt}$

Explanation:

The velocity-time graph is given below.

The area under curve will be (Trapezium Area)

$A=\frac{1}{2}(v_2+v_1)(t_2-t_1)$

Here we can see we are multiplying velocity with time, which will give displacement. 

So, the Area under the velocity-time graph gives the displacement.

Important Points

• Area under acceleration time graph gives velocity.
• The slope of the velocity-time graph gives acceleration.
• The slope of the distance-time graph gives speed.

Concept;

Equations of Motion

The equations of motion establish the relationship between acceleration, time, distance, initial speed, and final speed for a body moving in a straight line with uniform acceleration.

The equations are

1. v = u + at 
2. $s=ut+\frac{1}{2}a{t}^2$
3. v2 = u2 + 2as 

v is final velocity, u  is initial velocity, t is time, a is acceleration, s is the distance travelled.

Calculation:

Given:

v = 50 m/s, a = 2.5 m/s2?

v2 = u2 + 2as 

50² = 2 × 2.5 × s

s = 500 m

CONCEPT: 

• The equations of motion are used to characterize the movement of the body and are given by

$S=ut+\frac{1}{2}a{t}^2$

V = u + at

V² = U² + 2as

Where S = Displacement, t = time a =  acceleration, V = Final velocity, U = Initial velocity 

CALCULATION:

Given  S = 70 m, t = 7 sec u = 0m/s

• The equation of motion connecting displacement, initial velocity, and time is given by

$\Rightarrow S=ut+\frac{1}{2}a{t}^2$

substituting the given values the above equation becomes 

$\Rightarrow 70=0\times 7+\frac{1}{2}a\times 7^2$

$\Rightarrow a=\frac{70\times 2}{7\times 7}=\frac{20}{7}=2.85m/s^2$

• Hence, option 2 is the answer

The correct answer is velocity.

Key Points

• Acceleration is the rate at which velocity changes with respect to time.
• The ultimate outcome of all forces applied to a body, according to Newton's second law, is its acceleration. Acceleration, a vector quantity, determines the frequency at which a body's velocity varies. The formula can be used to express acceleration.
• The overall movement of an object, regardless of direction, is referred to as distance.
• The term "displacement" refers to a shift in an object's position. It's a vector quantity with a magnitude and a direction.
• The product of a particle's mass and velocity is called momentum. Momentum is a vector quantity in the sense that it has both a magnitude and a direction.

CONCEPT:

• Average speed: The total path length travelled divided by the total time interval during which the motion has taken place is called the average speed of the particle.

Total distance travelled by the object = 16 m + 16 m = 32 m

Total time taken = 4 s + 2 s = 6 s

$Averagespeed\left(\overline{v}\right)=\frac{totalpathlength\left(S\right)}{totaltimetaken\left(t\right)}=\frac{32}{6}=5.33m/s$

• Therefore, the average speed of the object is 5.33 m/s.

CONCEPT:

• Equation of Kinematics: These are the various relations between u, v, a, t, and s for the particle moving with uniform acceleration where the notations are used as:
• Equations of motion can be written as

V = U + at

$s=ut+\frac{1}{2}a{t}^2$

V2 =U2+ 2as

Where, U = Initial velocity, V = Final velocity, g= Acceleration due to gravity, t = time, and h= height/Distance covered

Where u = Initial velocity of the particle at time t = 0 sec

 v = Final velocity at time t sec

a = Acceleration of the particle

s = Distance travelled in time t sec

EXPLANATION:​

Given - Initial velocity (u) = 0 km/h, Final velocity (v) = 72 km/h = 20 m/s  and Time (t) = 5 min.

• As per the first law of motion

⇒ v = u + a t

⇒ 20 = 0 + (a₁ × 5 × 60)

$\Rightarrow a=\frac{20}{5\times 60}=\frac{1}{15}m{s}^2$

• The distance travelled by train 

$\Rightarrow S=ut+\left(\frac{1}{2}\times a\times t^2\right)$

$\Rightarrow S=0\times 300+\left(\frac{1}{2}\times \frac{1}{15}\times {300}^2\right)$

⇒ S = 3000 m = 3 km

• The distance travelled by train for attaining the above velocity is 3 km.

CONCEPT:

• Velocity: The rate of change of displacement of a body is called the velocity of that body.
  • Velocity is a vector quantity that has both magnitudes as well as direction.
• Acceleration: The rate of change of velocity is called the acceleration of the body.
  • Acceleration is also a vector quantity.
  • The slope of any velocity-time graph gives an acceleration of the body.
• Displacement: The minimum path length between two points is called displacement.
• Distance: The total path length between two points is called distance.

EXPLANATION:

• The slope of any graph is the ratio of the vertical change between two points to the horizontal change between the same points.
• In a velocity-time graph, the velocity (v) is present on the vertical axis while time (t) on the horizontal axis, so the slope of the graph is given by:

$\mathrm{S}\mathrm{l}\mathrm{o}\mathrm{p}\mathrm{e}=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}$

• Since the rate of change of velocity is termed as acceleration, the slope of a velocity-time graph gives the acceleration.
• Therefore the slope of the velocity-time graph for retarded motion is negative. Therefore option 3 is correct.

Additional Information

• Similarly, the slope of displacement - time graph gives velocity.
• The area under the velocity-time graph gives displacement and the area under the acceleration - time graph gives the change in velocity.

Formula Used:

Distance = Speed × Time

Calculations:

Let the speed of the train be ‘x’ m/s and the length of the train be ‘y’ m

A train crossed a 120 m long platform in 12 seconds,

⇒ 12 × (x) = 120 + y       ----(1)

A train crossed a 165 m long platform in 15 seconds,

⇒ 15 × (x) = 165 + y       ----(2)

Equation (2) - Equation (1) we get,

⇒ 3x = 45

⇒ x = 15 m/s = 15 × (18/5) km/h = 54 km/h