Given that a force`hatF` acts on a body for time t,and displaces the body by `hatd`.In which of the following cases,the speed of the body must not increase?

Options

A. `Fgtd`
B. `Fltd`
C. `hatF=hatd`
D. `hatF bot hatd`

Explanation

To solve the problem, we will analyze the situation using the work-energy theorem. Here’s a step-by-step breakdown:

### Step 1: Understand the Work-Energy Theorem
The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. Mathematically, this is expressed as:

\[ W = \Delta KE \]

Where \( W \) is the work done on the body and \( \Delta KE \) is the change in kinetic energy.

### Step 2: Determine Conditions for No Change in Speed
For the speed of the body to remain constant, the change in kinetic energy must be zero:

\[ \Delta KE = 0 \]

This implies that the work done on the body must also be zero:

\[ W = 0 \]

### Step 3: Express Work Done in Terms of Force and Displacement
The work done by a force on an object is given by the dot product of the force vector and the displacement vector:

\[ W = \vec{F} \cdot \vec{d} = F d \cos \theta \]

Where:
- \( \vec{F} \) is the force vector,
- \( \vec{d} \) is the displacement vector,
- \( \theta \) is the angle between the force and displacement vectors.

### Step 4: Analyze the Condition for Zero Work Done
For the work done \( W \) to be zero, the cosine of the angle \( \theta \) must be zero:

\[ F d \cos \theta = 0 \]

This occurs when \( \cos \theta = 0 \), which happens when:

\[ \theta = 90^\circ \]

This means that the force is acting perpendicular to the direction of displacement.

### Step 5: Conclusion
Thus, the speed of the body must not increase when the force acting on it is perpendicular to the displacement. Therefore, the correct answer is when the force vector is perpendicular to the displacement vector.

### Final Answer
The speed of the body must not increase when the force vector is perpendicular to the displacement vector.

---