If `u = 12m//s, a =-3m//s^(2)` find t at which v = 0 and displacement between 2s and 5s

Options

A. 1s, 3m
B. 2s, 1m
C. 4s, 4m
D. 4s, 4.5m

Answer

To solve the problem step by step, we will first find the time \( t \) at which the velocity \( v \) becomes zero, and then we will calculate the displacement between \( t = 2 \) seconds and \( t = 5 \) seconds.

### Step 1: Find the time \( t \) when \( v = 0 \)

We use the equation of motion:
\[ v = u + at \]
Where:
- \( u = 12 \, \text{m/s} \) (initial velocity)
- \( a = -3 \, \text{m/s}^2 \) (acceleration)

Setting \( v = 0 \):
\[ 0 = 12 - 3t \]

Now, we can solve for \( t \):
\[ 3t = 12 \]
\[ t = \frac{12}{3} = 4 \, \text{s} \]

### Step 2: Calculate the displacement at \( t = 5 \) seconds

We use the equation for displacement:
\[ s = ut + \frac{1}{2} a t^2 \]

Substituting \( t = 5 \) seconds:
\[ s_5 = 12 \times 5 + \frac{1}{2} \times (-3) \times (5^2) \]
Calculating each term:
\[ s_5 = 60 - \frac{1}{2} \times 3 \times 25 \]
\[ s_5 = 60 - \frac{75}{2} \]
\[ s_5 = 60 - 37.5 = 22.5 \, \text{m} \]

### Step 3: Calculate the displacement at \( t = 2 \) seconds

Now, we calculate the displacement at \( t = 2 \) seconds:
\[ s_2 = 12 \times 2 + \frac{1}{2} \times (-3) \times (2^2) \]
Calculating each term:
\[ s_2 = 24 - \frac{1}{2} \times 3 \times 4 \]
\[ s_2 = 24 - 6 = 18 \, \text{m} \]

### Step 4: Calculate the displacement between \( t = 2 \) seconds and \( t = 5 \) seconds

The displacement between \( t = 2 \) seconds and \( t = 5 \) seconds is given by:
\[ \Delta s = s_5 - s_2 \]
Substituting the values we found:
\[ \Delta s = 22.5 - 18 = 4.5 \, \text{m} \]

### Final Answers

- The time \( t \) at which \( v = 0 \) is \( 4 \, \text{s} \).
- The displacement between \( t = 2 \) seconds and \( t = 5 \) seconds is \( 4.5 \, \text{m} \).