#### Position of an ant (S in metres) moving in Y-Z plane is given by $S= 2t^2 \hat j + 5 \hat k$ (where t is in second). The magnitude and direction of velocity of the ant at t = 1 s will be :

(A) 16 m/s in y-direction

(B) 4 m/s in x-direction

(C) 9 m/s in z-direction

(D) 4 m/s in y-direction

### Solution:

This question was asked in **JEE-Mains 2024** on January 27, 2024 in the morning session.

**Required Utensils:**

- Vectors
- Differentiation (Click to know formulas of differentiation)
- Basic Differentiation formula $\frac{d}{dt}(t^n) = nt^{n-1}$

**Preparation:**

This is a simple differentiation question. You will find questions like this in most of the physics

Here $S= 2t^2 \hat j + 5 \hat k$

so $v = \frac{dS}{dt} = 4t \hat j$

Since $5 \hat k$ is a constant because there is no term of $t$ here.

Substituting $t = 1$

$v=4 \times 1 \hat j$

$v = 4 \hat j$

Since $\hat j$ is there, so velocity is in +Y direction with magnitude 4.

Hence **option (D)** is the correct answer

### Similar Questions

**An object moves along the X-axis according to the equation $x = 3t^3 – 2t + 4$. What is the acceleration of the object at $t = 2$ s?**

A) 34 m/s²

B) 36 m/s²

C) 38 m/s²

D) 40 m/s²

2. **A particle’s position is given by $r = (4t \hat i + 2t^2 \hat j + 3 \hat k)$, where $t$ is in seconds. What is the velocity of the particle at $t = 3$ s?**

A) $4 \hat i + 12 \hat j$ m/s

B) $12 \hat i + 6 \hat j$ m/s

C) $4 \hat i + 6 \hat j$ m/s

D) $12 \hat i + 12 \hat j$ m/s

3. **A car accelerates from rest with a constant acceleration of $5 m/s^2$ in the positive Y-direction. What is the velocity of the car after 4 seconds?**

A) 15 m/s

B) 20 m/s

C) 25 m/s

D) 30 m/s

**4. A projectile is launched with an initial velocity of $v_0 = 20 \hat i + 10 \hat j$ m/s. What is the maximum height reached by the projectile?**

A) 5 m

B) 10 m

C) 15 m

D) 20 m

### Resources:

- Differentiation
- Sarthak Solution
- Video Solution

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