14. A car starts from rest and accelerates at $$ 1 \mathrm{~ms}^{-2} $$ for 10 seconds. It then continues at a uniform speed for another 20 seconds and decelerates to rest in 5 seconds. Find the: (a) distance travelled; (b) average speed; (c) time taken to cover half the distance.

Question

UpStudy AI · Accepted Answer

To solve the problem, we will break it down into three parts based on the car's motion: acceleration, uniform speed, and deceleration.

### Known Conditions:
1. **Acceleration phase:**
   - Initial speed $$ u = 0 \, \text{m/s} $$
   - Acceleration $$ a = 1 \, \text{m/s}^2 $$
   - Time $$ t_1 = 10 \, \text{s} $$

2. **Uniform speed phase:**
   - Time $$ t_2 = 20 \, \text{s} $$

3. **Deceleration phase:**
   - Time $$ t_3 = 5 \, \text{s} $$
   - Final speed $$ v = 0 \, \text{m/s} $$

### Step-by-Step Solution:

#### (a) Distance travelled

1. **Distance during acceleration:**
   The distance $$ s_1 $$ during the acceleration phase can be calculated using the formula:
   $$
   s_1 = ut + \frac{1}{2} a t^2
   $$
   Substituting the known values:
   $$
   s_1 = 0 \cdot 10 + \frac{1}{2} \cdot 1 \cdot (10)^2 = \frac{1}{2} \cdot 1 \cdot 100 = 50 \, \text{m}
   $$

2. **Distance during uniform speed:**
   The speed at the end of the acceleration phase is given by:
   $$
   v = u + at = 0 + 1 \cdot 10 = 10 \, \text{m/s}
   $$
   The distance $$ s_2 $$ during the uniform speed phase is:
   $$
   s_2 = v \cdot t_2 = 10 \cdot 20 = 200 \, \text{m}
   $$

3. **Distance during deceleration:**
   The initial speed for the deceleration phase is $$ v = 10 \, \text{m/s} $$ and the final speed is $$ 0 \, \text{m/s} $$. The average speed during deceleration is:
   $$
   v_{\text{avg}} = \frac{v + 0}{2} = \frac{10 + 0}{2} = 5 \, \text{m/s}
   $$
   The distance $$ s_3 $$ during the deceleration phase is:
   $$
   s_3 = v_{\text{avg}} \cdot t_3 = 5 \cdot 5 = 25 \, \text{m}
   $$

4. **Total distance travelled:**
   $$
   s_{\text{total}} = s_1 + s_2 + s_3 = 50 + 200 + 25 = 275 \, \text{m}
   $$

#### (b) Average speed

The total time taken $$ T $$ is:
$$
T = t_1 + t_2 + t_3 = 10 + 20 + 5 = 35 \, \text{s}
$$
The average speed $$ v_{\text{avg}} $$ is given by:
$$
v_{\text{avg}} = \frac{s_{\text{total}}}{T} = \frac{275}{35} \approx 7.857 \, \text{m/s}
$$

#### (c) Time taken to cover half the distance

Half the total distance is:
$$
\frac{s_{\text{total}}}{2} = \frac{275}{2} = 137.5 \, \text{m}
$$

1. **Distance covered in the acceleration phase:**
   - The car covers $$ 50 \, \text{m} $$ in the first $$ 10 \, \text{s} $$.

2. **Distance covered in the uniform speed phase:**
   - The car covers $$ 200 \, \text{m} $$ in the next $$ 20 \, \text{s} $$, reaching a total of $$ 250 \, \text{m} $$ by the end of this phase.

Since $$ 137.5 \, \text{m} $$ is less than $$ 250 \, \text{m} $$, we need to find the time taken to cover the remaining distance after the acceleration phase.

3. **Remaining distance after acceleration:**
   $$
   \text{Remaining distance} = 137.5 - 50 = 87.5 \, \text{m}
   $$

4. **Time taken to cover $$ 87.5 \, \text{m} $$ at $$ 10 \, \text{m/s} $$:**
   $$
   t = \frac{\text{distance}}{\text{speed}} = \frac{87.5}{10} = 8.75 \, \text{s}
   $$

5. **Total time to cover half the distance:**
   $$
   \text{Total time} = t_1 + t = 10 + 8.75 = 18.75 \, \text{s}
   $$

### Final Answers:
(a) Distance travelled: $$ 275 \, \text{m} $$  
(b) Average speed: $$ \approx 7.857 \, \text{m/s} $$  
(c) Time taken to cover half the distance: $$ 18.75 \, \text{s} $$
(a) Distance traveled: 275 meters (b) Average speed: approximately 7.86 m/s (c) Time to cover half the distance: 18.75 seconds

14. A car starts from rest and accelerates at \( 1 \mathrm{~ms}^{-2} \) for 10 seconds. It then continues at a uniform speed for another 20 seconds and decelerates to rest in 5 seconds. Find the: (a) distance travelled; (b) average speed; (c) time taken to cover half the distance.

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Physics Questions