Question

A T. Rex is chasing a wounded Triceratops that is running at a rate of 9.94 m/s. The T. Rex is initially running at 3.2 m/s. How long does the T. Rex need to accelerate at a rate of 3.47m/s2 for it to be able to catch his prey. G/U E S S

Answer 1

To determine how long the T. Rex needs to accelerate in order to catch the wounded Triceratops, we can set up an equation based on the position of both dinosaurs.

Let:
- \( v_t = 3.2 \, \text{m/s} \) (initial speed of T. Rex)
- \( a_t = 3.47 \, \text{m/s}^2 \) (acceleration of T. Rex)
- \( v_c = 9.94 \, \text{m/s} \) (constant speed of Triceratops)
- \( t \) = time in seconds

### Step 1: Position equations
1. **Position of the Triceratops (which is moving at a constant speed)**:
\[
d_c = v_c \cdot t = 9.94 t
\]

2. **Position of the T. Rex (which is accelerating)**:
The position of an object under constant acceleration can be calculated using the equation:
\[
d_t = v_t \cdot t + \frac{1}{2} a_t t^2 = 3.2 t + \frac{1}{2} (3.47) t^2
\]
Simplifying the T. Rex position equation:
\[
d_t = 3.2 t + 1.735 t^2
\]

### Step 2: Set the distances equal to each other
To find the time \( t \) when the T. Rex catches the Triceratops, we set their positions equal:
\[
9.94 t = 3.2 t + 1.735 t^2
\]

### Step 3: Rearranging the equation
Rearranging gives us:
\[
1.735 t^2 + 3.2 t - 9.94 t = 0
\]
\[
1.735 t^2 - 6.74 t = 0
\]

### Step 4: Factor out t
Factoring gives:
\[
t(1.735 t - 6.74) = 0
\]

This gives us two solutions:
1. \( t = 0 \) (initial time)
2. \( 1.735 t - 6.74 = 0 \)

### Step 5: Solve for \( t \)
Solving the second equation:
\[
1.735 t = 6.74
\]
\[
t = \frac{6.74}{1.735} \approx 3.88 \, \text{s}
\]

### Conclusion
The T. Rex needs approximately **3.88 seconds** to accelerate at a rate of 3.47 m/s² in order to catch the wounded Triceratops.