If acceleration of a particle at any time is given by `a = 2t + 5`
Calculate the velocity after 5 s, if it starts from rest

Options

A. `50 ms ^(-1)`
B. `25 ms ^(-1)`
C. `100 ms ^(-1)`
D. `75ms^(-1)`

Answer

To solve the problem of finding the velocity of a particle after 5 seconds, given its acceleration as a function of time, we can follow these steps:

### Step 1: Understand the relationship between acceleration and velocity
Acceleration \( a \) is defined as the rate of change of velocity with respect to time, which can be mathematically expressed as:
\[ a = \frac{dv}{dt} \]
This means that the change in velocity \( dv \) can be expressed as:
\[ dv = a \, dt \]

### Step 2: Substitute the given acceleration function
The acceleration of the particle is given as:
\[ a = 2t + 5 \]
Substituting this into the equation for \( dv \):
\[ dv = (2t + 5) \, dt \]

### Step 3: Integrate to find the velocity
To find the velocity, we need to integrate both sides. The limits for the velocity will be from the initial velocity \( v_0 \) (which is 0 since the particle starts from rest) to the final velocity \( v \) after 5 seconds. The limits for time will be from 0 to 5 seconds:
\[ \int_{0}^{v} dv = \int_{0}^{5} (2t + 5) \, dt \]

### Step 4: Calculate the integral
The left side integrates to:
\[ v - 0 = v \]
Now, we calculate the right side:
\[ \int_{0}^{5} (2t + 5) \, dt = \int_{0}^{5} 2t \, dt + \int_{0}^{5} 5 \, dt \]
Calculating each integral separately:
1. For \( \int 2t \, dt \):
\[ \int 2t \, dt = t^2 \bigg|_{0}^{5} = 5^2 - 0^2 = 25 \]
2. For \( \int 5 \, dt \):
\[ \int 5 \, dt = 5t \bigg|_{0}^{5} = 5 \cdot 5 - 5 \cdot 0 = 25 \]

Adding these results together:
\[ \int_{0}^{5} (2t + 5) \, dt = 25 + 25 = 50 \]

### Step 5: Set the equation for velocity
Now we have:
\[ v = 50 \]

### Final Result
The velocity of the particle after 5 seconds is:
\[ \boxed{50 \, \text{m/s}} \]