The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by…

Q60 jee_main_2024_01_february_morning Kinematics

A particle is moving in one dimension (along x-axis) under the action of a variable force. It's initial position was

16~m

right of origin. The variation of its position (

x

) with time (

t

) is given as

x = -3t³ + 18t² + 16t

where

x

is in m and

t

is in s. The velocity of the particle when its acceleration becomes zero is _______

m/s

.

Numerical Answer. Answer: 52 to 52

Solution

Related Formula

Velocity via first derivative of position:

v = (dx)/(dt)

Acceleration via derivative of velocity:

a = (dv)/(dt) = (d²x)/(dt²)

Core Logic

Given $x(t) = -3t³ + 18t² + 16t$ $x(t) = -3t^3 + 18t^2 + 16t$. Differentiate once to find velocity $v(t)$ $v(t)$:

v = (dx)/(dt) = -9t² + 36t + 16

Differentiate again to find acceleration $a(t)$ $a(t)$:

a = (dv)/(dt) = -18t + 36

Step 1: Find the Time When Acceleration is Zero

Set the acceleration equation to zero:

-18t + 36 = 0 18t = 36 t = 2~s

Step 2: Calculate Velocity at Target Time

Substitute $t = 2~s$ $t = 2\mathrm{~s}$ back into the velocity function:

v(2) = -9(2)² + 36(2) + 16

v(2) = -36 + 72 + 16 = 36 + 16 = 52~ms⁻¹

Note on original structural text step matching: The original calculation step text features a minor print truncation ( $v = -9t² + 36 + 16$ $v = -9t^2 + 36 + 16$), but explicitly evaluates out to the correct total of $52~m/s$ $52\mathrm{~m/s}$.

Pattern Recognition

Inflection point tracking: Finding where acceleration equals zero is mathematically identical to finding the maximum velocity point on the curve.

Chapter Mix

Class 11 Physics: Motion in a Straight Line

Q47 jee_main_2024_29_january_evening Kinematics of Linear Motion

A particle is moving in a straight line. The variation of position

x

as a function of time

t

is given as

x = (t³ - 6t² + 20t + 15) m

. The velocity of the body when its acceleration becomes zero is:

A. $4 m/s$ $4\text{ m/s}$
B. $8 m/s$ $8\text{ m/s}$
C. $10 m/s$ $10\text{ m/s}$
D. $6 m/s$ $6\text{ m/s}$

Solution

Related Formula

The relationship between position $x$ $x$, velocity $v$ $v$, and acceleration $a$ $a$ is given by differentiation with respect to time $t$ $t$:

v = (dx)/(dt)

a = (dv)/(dt)

Core Logic

Given position:

x = t³ - 6t² + 20t + 15

Differentiating once to find velocity $v$ $v$:

v = (dx)/(dt) = 3t² - 12t + 20

Differentiating again to find acceleration $a$ $a$:

a = (dv)/(dt) = 6t - 12

Step 1: Determine Time when Acceleration is Zero

Set the acceleration to zero:

a = 0 6t - 12 = 0 t = 2 seconds

So, the acceleration becomes zero at $t = 2 s$ $t = 2\text{ s}$.

Step 2: Calculate Velocity at this Time

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

v = 3(2)² - 12(2) + 20

v = 3(4) - 24 + 20

v = 12 - 24 + 20 = 8 m/s

Thus, the velocity is $8 m/s$ $8\text{ m/s}$.

Pattern Recognition

Sees: displacement function of degree 3 → acceleration is linear in time. The zero of a linear function of form $At - B = 0$ $At - B = 0$ is easily calculated, and substituting back resolves to a basic quadratic.

Chapter Mix

Class 11 Physics: Motion in a Straight Line

Q31 jee_main_2024_27_jan_morning Velocity and Acceleration in Two Dimensions

Position of an ant (

S

in metres) moving in Y-Z plane is given by

S = 2t² j + 5 k

(where

t

is in seconds). The magnitude and direction of velocity of the ant at

t = 1

s will be:

A. $16 m/s in y-direction$ $16\text{ m/s in y-direction}$
B. $4 m/s in x-direction$ $4\text{ m/s in x-direction}$
C. $9 m/s in z-direction$ $9\text{ m/s in z-direction}$
D. $4 m/s in y-direction$ $4\text{ m/s in y-direction}$

Solution

Related Formula

v = d Sdt

Core Logic

Differentiating the position vector with respect to time gives the velocity vector:

v = (d)/(dt)(2t² j + 5 k) = 4t j

Step 1: Evaluation at given time

Substitute $t = 1 s$ $t = 1\text{ s}$ into the velocity expression:

v = 4(1) j = 4 j m/s

Pattern Recognition

Constant term vectors (like $5 k$ $5\hat{k}$) drop out during differentiation. The resulting velocity only has a component along the $j$ $\hat{j}$ direction, confirming motion purely parallel to the y-axis at that instant.

Chapter Mix

Class 11 Physics: Kinematics

Q51 jee_main_2024_27_jan_morning Motion in a Plane with Constant Acceleration

A particle starts from the origin at

t = 0

with a velocity

5 i m/s

and moves in the x-y plane under the action of a force which produces a constant acceleration of

(3 i + 2 j) m/s²

. If the x-coordinate of the particle at that instant is

84 m

, then the speed of the particle at this time is

\sqrt(a) m/s

. The value of

a

is ______.

Numerical Answer. Answer: 673 to 673

Solution

Related Formula

vₓ² - uₓ² = 2 aₓ x

vy = uy + ay t

Core Logic

Analyze the motion along the x-axis first to find the final x-velocity component ( $uₓ = 5 m/s$ $u_x = 5\text{ m/s}$, $aₓ = 3 m/s²$ $a_x = 3\text{ m/s}^2$, $x = 84 m$ $x = 84\text{ m}$):

vₓ² - 5² = 2(3)(84)

vₓ² - 25 = 504 vₓ² = 529 vₓ = 23 m/s

Step 1: Compute time interval

Using the velocity relation along the x-axis:

vₓ = uₓ + aₓ t 23 = 5 + 3t 3t = 18 t = 6 s

Step 2: Evaluate y-axis velocity component

Since $uy = 0$ $u_y = 0$ and $ay = 2 m/s²$ $a_y = 2\text{ m/s}^2$, calculate $vy$ $v_y$ at $t = 6 s$ $t = 6\text{ s}$:

vy = 0 + 2(6) = 12 m/s

Step 3: Total net speed calculation

v² = vₓ² + vy² = 23² + 12² = 529 + 144 = 673

v = \sqrt(673) m/s

Comparing this with $\sqrt(a)$ $\sqrt{a}$ yields $a = 673$ $a = 673$.

Pattern Recognition

Splitting coordinates explicitly isolates vector calculation lines cleanly, optimizing equation selections relative to independent components.

Chapter Mix

Class 11 Physics: Kinematics

Q jee_main_2024_29_jan_morning Projectile Motion

A ball rolls off the top of a stairway with horizontal velocity

u

. The steps are

0.1 ~m

high and

0.1 ~m

wide. The minimum velocity

u

with which that ball just hits the step 5 of the stairway will be

\sqrt(x) ~ms⁻¹

where

x =

________ [use

g = 10 ~m/s²

].

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

For a projectile fired horizontally from height $h$ $h$ with speed $u$ $u$:

Horizontal Range R = u \cdot t

Vertical Displacement h = (1)/(2) g t²

Core Logic

To just clear step 4 and land on the tread of step 5, the flight trajectory must pass beyond the outer corner boundary vertex of the $4th$ $4^{\text{th}}$ step.

Therefore, net horizontal distance required to cross 4 complete steps is:

R = 4 \times 0.1 ~m = 0.4 ~m

Similarly, net vertical fall matching the top of the $4th$ $4^{\text{th}}$ step level is:

h = 4 \times 0.1 ~m = 0.4 ~m

Trajectory diagram of a ball clearing step corners on a staircase grid for Q52

Step 1: Determine Time of Flight

Using the vertical kinematic equation:

0.4 = (1)/(2) \times 10 \times t² 0.4 = 5 t²

t² = (0.4)/(5) = 0.08 ~s²

Step 2: Determine Velocity

Using the horizontal path equation:

R = u \cdot t R² = u² \cdot t²

Substituting range and time values:

(0.4)² = u² \times 0.08

0.16 = u² \times 0.08

u² = (0.16)/(0.08) = 2 u = \sqrt(2) ~m/s

Step 3: Extract x

Matching the parameter form $u = \sqrt(x)$ $u = \sqrt{x}$, we find:

$x = 2$ $x = 2$

Pattern Recognition

To clear the $n$ $n$-th step, the projectile must safely pass the outer point of step $(n-1)$ $(n-1)$. Treat the geometric coordinates of corners as bounding constraints ( $x = (n-1)w$ $x = (n-1)w$, $y = (n-1)h$ $y = (n-1)h$) to configure kinematics instantaneously.

Chapter Mix

Class 11 Physics: Motion in a Plane

The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between t = 0$t = 0$ to t = 4s$t = 4s$ ? The velocity-time profile tracking motion across consecutive geometric shapes up to 4 seconds.

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Motion in a Straight Line Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Find the Time When Acceleration is Zero

Step 2: Calculate Velocity at Target Time

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine Time when Acceleration is Zero

Step 2: Calculate Velocity at this Time

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluation at given time

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute time interval

Step 2: Evaluate y-axis velocity component

Step 3: Total net speed calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine Time of Flight

Step 2: Determine Velocity

Step 3: Extract x

Pattern Recognition

Chapter Mix

More Motion in a Straight Line Questions — jee_main_2025_28_jan_evening

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The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between $t = 0$ $t = 0$ to $t = 4s$ $t = 4s$ ?
The velocity-time profile tracking motion across consecutive geometric shapes up to 4 seconds.