A car starts from rest and moves with constant acceleration along a straight horizontal road. The car reaches a speed of $V \, m \, s^{-1}$ in 20 seconds. It moves at constant speed $V \, m \, s^{-1}$ for the next 30 seconds, then moves with constant deceleration $\frac{1}{2} \, m \, s^{-2}$ until it has speed 8 m s$^{-1}$. It moves at speed 8 m s$^{-1}$ for the next 15 seconds and then moves with constant deceleration $\frac{1}{3} \, m \, s^{-2}$ until it comes to rest. In the first 20 seconds of this journey the car travels 140 m. Find (b) the value of $V$. (c) the total time for this journey, (d) the total distance travelled by the car.

Photo AI

A car starts from rest and moves with constant acceleration along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 3 - 2014 - Paper 1

Question 3

A car starts from rest and moves with constant acceleration along a straight horizontal road. The car reaches a speed of $V \, m \, s^{-1}$ in 20 seconds. It moves a... show full transcript

Worked Solution & Example Answer:A car starts from rest and moves with constant acceleration along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 3 - 2014 - Paper 1

Step 1

(b) the value of V

96%

114 rated

Answer

To find the value of $V$ , we can use the distance traveled during the first 20 seconds. The formula for distance with constant acceleration is:

$s = ut + \frac{1}{2} a t^2$

Here, the initial speed $u = 0$ , acceleration $a = \frac{V}{20}$ (since it reaches $V$ in 20 seconds), and time $t = 20$ seconds. Plugging these values in gives:

$140 = 0 \cdot 20 + \frac{1}{2} \left( \frac{V}{20} \right) (20)^2$

This simplifies to:

$140 = \frac{1}{2} \cdot V \cdot 20$

From this, we can isolate $V$ :

$$140 = 10V \Rightarrow V = 14 , m , s^{-1}.$

Step 2

(c) the total time for this journey

99%

104 rated

Answer

To determine the total time, we break the journey into segments:

Acceleration Phase:
- Time taken to reach speed $V$ is 20 seconds.
Constant Speed Phase:
- Time at constant speed $V = 14 \, m \, s^{-1}$ for 30 seconds.
Deceleration to 8 m s $^{-1}$ Phase:
- Using the formula $v = u + at$ $v = u + a t$ , we find:
  - $8 = 14 - \frac{1}{2}(t_1)$
  - $t_1 = 12$ seconds (for deceleration to 8 m s $^{-1}$ ).
Constant Speed Phase at 8 m s $^{-1}$ :
- This lasts for 15 seconds.
Final Deceleration Phase to Rest:
- The speed decreases from 8 m s $^{-1}$ $^{- 1}$ to rest with deceleration $\frac{1}{3} \, m \, s^{-2}$ $\frac{1}{3} m s^{- 2}$ :
  - Using $0 = 8 - \frac{1}{3} t_2$ gives:
  - $t_2 = 24$ seconds.

Adding all of these together:

$$Total \ Time = 20 + 30 + 12 + 15 + 24 = 101 seconds.$

Step 3

(d) the total distance travelled by the car

96%

101 rated

Answer

The total distance can be calculated by summing the distances of each segment of the journey:

Acceleration Phase:
- We already found this distance as 140 m.
Constant Speed Phase:
- Distance $s = vt = 14 \, m \, s^{-1} \times 30 \, s = 420 \, m.$
Constant Speed at 8 m s $^{-1}$ Phase:
- Distance $s = vt = 8 \, m \, s^{-1} \times 15 \, s = 120 \, m.$
Final Deceleration Phase:
- Using average speed:
- Average speed = $\frac{(8 + 0)}{2} = 4 \, m \, s^{-1}$ .
- Distance $s = vt = 4 \, m \, s^{-1} \times 24 \, s = 96 \, m.$

Now, summing up all distances:

$Total \ Distance = 140 + 420 + 120 + 96 = 776 \, m.$

A car starts from rest and moves with constant acceleration along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 3 - 2014 - Paper 1

Worked Solution & Example Answer:A car starts from rest and moves with constant acceleration along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 3 - 2014 - Paper 1

(b) the value of V

(c) the total time for this journey

(d) the total distance travelled by the car

Join the A-Level students using SimpleStudy...