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 Type:
Enter a numerical value Answer: 52 to 52 +4 marks

Solution & Explanation

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. Differentiate once to find velocity v(t):

v = (dx)/(dt) = -9t² + 36t + 16

Differentiate again to find acceleration 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 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), but explicitly evaluates out to the correct total of 52~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

Reference Study Guides

Motion In A Straight Line

More Motion in a Straight Line Previous-Year Questions — Page 5

Q8 jee_main_2025_24_jan_evening Velocity and Acceleration
The position vector of a moving body at any instant of time is given as r = (5t² i - 5t j)m. The magnitude and direction of velocity at t = 2s is,
  • A. 5√(15) m/s, making an angle of ⁻¹4 with -ve Y axis
  • B. 5√(15) m/s, making an angle of ⁻¹4 with +ve X axis
  • C. 5√(17) m/s, making an angle of ⁻¹4 with -ve Y axis
  • D. 5√(17) m/s, making an angle of ⁻¹4 with +ve X axis

Solution

Related Formula
v = d rdt
Core Logic

Given position vector:

r = 5t² i - 5t j

Differentiating with respect to t:

v = 10t i - 5 j

At t = 2 s:

v = 20 i - 5 j

Magnitude of velocity:

v = √((20)² + (-5)²) = √(400 + 25) = √(425) = 5√(17) m/s

Direction analysis:

Velocity vector components angle calculation Q8
Velocity vector components angle calculation Q8
vₓ = 20 along +x axis, vy = -5 along -y axis. Let θ be the angle made with the negative Y-axis:

θ = |(vₓ)/(vy)| = (20)/(5) = 4 θ = ⁻¹4
Pattern Recognition

Always read the reference axis carefully in direction questions. Here, the angle is measured from the negative Y-axis, making θ = vₓ / |vy|.

Chapter Mix

Class 11 Physics: Motion in a Plane

Q9 jee_main_2025_28_jan_evening Velocity Time Graphs
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 to t = 4s ?
Velocity Time Graphs diagram for Q9 - JEE Main 2025 Evening
The velocity-time profile tracking motion across consecutive geometric shapes up to 4 seconds.
  • A. 30m
  • B. 10m
  • C. 13m
  • D. 11m

Solution

Related Formula

The distance traveled by an object equals the total area enclosed under its velocity-time (v-t) plot along the time axis, treating all regional boundaries as strictly positive metrics:

Distance = ∫ |v| dt
Core Logic

From the geometric grid profile between t = 0 and t = 4 s:

  • First Region (t=0 to t=2 s): Forms a \right-angled \triangle with base = 2 s and peak height = 10 ms⁻¹.
Area₁ = (1)/(2) × 2 × 10 = 10 m
  • Second Region (t=2 to t=4 s): Forms a standard rectangle with width = (4 - 2) = 2 s and height = 10 ms⁻¹.
Area₂ = 2 × 10 = 20 m

Summing the areas together to extract total displacement path:

Total Distance = 10 + 20 = 30 m
Pattern Recognition

Always differentiate between distance and displacement on graph tracks. Displacement treats components below the axis as negative fields, while distance calculates absolute geometric magnitudes without direction.

Chapter Mix

Class 11 Physics: Motion in a Straight Line

Q jee_main_2025_29_jan_morning Projectile Motion
Two projectiles are fired with same initial speed from same point on ground at angles of (45° - α) and (45° + α) , respectively, with the horizontal direction. The ratio of their maximum heights attained is :
  • A. (1 - α)/(1 + α)
  • B. (1 + α)/(1 - α)
  • C. (1 - 2α)/(1 + 2α)
  • D. (1 + 2α)/(1 - 2α)

Solution

Related Formula
H = (u² ² θ)/(2g)
Core Logic

Let θ₁ = 45° - α and θ₂ = 45° + α. The ratio of maximum heights is:

(H₁)/(H₂) = ²(45° - α) ²(45° + α)
Step 1: Simplify Trigonometric Terms
(45° ± α) = 1√(2) α ± 1√(2) α (H₁)/(H₂) = (( α - α)²)/(( α + α)²) = ( ²α + ²α - 2 α α)/( ²α + ²α + 2 α α) = (1 - 2α)/(1 + 2α)
Pattern Recognition

For complementary angles shifted symmetric to 45°, ratios simplify cleanly via 2α identities.

Chapter Mix

Class 11 Physics: Motion in a Plane

Q jee_main_2025_29_jan_morning Relative Motion
The maximum speed of a boat in still water is 27 km/h . Now this boat is moving downstream in a river flowing at 9 km/h . A man in the boat throws a ball vertically upwards with speed of 10 m/s . Range of the ball as observed by an observer at rest on the river bank, is ________ cm. (Take g = 10 m/s² )
Numerical Answer. Answer: 2000 to 2000

Solution

Related Formula
T = (2uy)/(g), R = vₓ · T
Core Logic

Relative range tracking frame vector
Relative range tracking frame vector

The total horizontal velocity components combine due to downstream motion addition

vₓ = 27 + 9 = 36 km/h = 36 · (5)/(18) = 10 m/s

The time of flight for the vertical launch tracking stands at :

T = (2 · 10)/(10) = 2 s

Horizontal range measured along the river bank frame is :

Range = vₓ · T = 10 · 2 = 20 m = 2000 cm
Chapter Mix

Class 11 Physics: Motion in a Straight Line Class 11 Physics: Motion in a Plane

Q47 jee_main_2024_01_february_morning Projectile Motion
A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle θ to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection θ is then given by:
  • A. ⁻¹[ 2gT²pi²R](1)/(2)
  • B. ⁻¹[ π²R2gT²](1)/(2)
  • C. ⁻¹[ 2gT²pi²R](1)/(2)
  • D. ⁻¹[(π R)/(2gT²)](1)/(2)

Solution

Related Formula

Uniform circular velocity:

v = (2π R)/(T)

Maximum height of a projectile:

H = (v² ²θ)/(2g)
Core Logic

Given that the projectile max height matches H = 4R:

4R = (v² ²θ)/(2g)

Substitute the value of v from the circular motion loop:

4R = (((2π R)/(T))² ²θ)/(2g) 4R = (4π² R² ²θ)/(2g T²)
Step 1: Isolate Angular Components

Cancel out 4R from both sides:

1 = (π² R ²θ)/(2g T²) ²θ = (2g T²)/(π² R) θ = ((2g T²)/(π² R))(1)/(2) θ = ⁻¹[(2gT²)/(π² R)](1)/(2)
Pattern Recognition

Connect circular metrics directly to projectile parameters via velocity matching. Keeping terms unsimplified makes it easy to cancel common elements later.

Chapter Mix

Class 11 Physics: Motion in a Plane

More Motion in a Straight Line Questions — jee_main_2024_01_february_morning

Practice all Motion in a Straight Line previous-year questions →

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