Question

A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

x = Question Blank 1 of 6
Answer

Vi = Question Blank 2 of 6
Answer

Vf = Question Blank 3 of 6
Answer

a = Question Blank 4 of 6
Answer

t = Question Blank 5 of 6
Answer

Answer 1

To determine the acceleration of the car, we can use the kinematic equation for uniformly accelerated motion:

\[
x = V_i t + \frac{1}{2} a t^2
\]

where:
- \(x\) is the distance traveled (110 m),
- \(V_i\) is the initial velocity (0 m/s, since the car starts from rest),
- \(a\) is the acceleration (unknown),
- \(t\) is the time (5.21 s).

Since the car starts from rest, \(V_i = 0\), the equation simplifies to:

\[
x = \frac{1}{2} a t^2
\]

Now plug in the values:

\[
110 = \frac{1}{2} a (5.21)^2
\]

Calculating \( (5.21)^2 \):

\[
(5.21)^2 = 27.1441
\]

So the equation becomes:

\[
110 = \frac{1}{2} a (27.1441)
\]

Multiply both sides by 2 to eliminate the fraction:

\[
220 = a \cdot 27.1441
\]

Now, solve for \(a\):

\[
a = \frac{220}{27.1441} \approx 8.11 \, \text{m/s}^2
\]

Now we have determined the acceleration. Let's fill in the answers:

1. **Acceleration (a)**: \(8.11 \, \text{m/s}^2\)

2. **Initial Velocity (Vi)**: Since the car starts from rest, it is \(0 \, \text{m/s}\).

3. **Final Velocity (Vf)**: We can find the final velocity using the equation:

\[
V_f = V_i + a t
\]

Substituting in the values we know:

\[
V_f = 0 + (8.11)(5.21) \approx 42.28 \, \text{m/s}
\]

4. **Distance (x)**: \(110 \, \text{m}\)

5. **Time (t)**: \(5.21 \, \text{s}\)

### Summary of the Answers

- **Question Blank 1 (x)**: \(110 \, \text{m}\)
- **Question Blank 2 (Vi)**: \(0 \, \text{m/s}\)
- **Question Blank 3 (Vf)**: \(42.28 \, \text{m/s}\)
- **Question Blank 4 (a)**: \(8.11 \, \text{m/s}^2\)
- **Question Blank 5 (t)**: \(5.21 \, \text{s}\)