A cyclist travels in a straight line so that his
displacement, metres, from a fixed point , is
given by , where is the time in
seconds after passing a point on the line.
(i) Find the distance .
(ii) Find the velocity of the cyclist at .
(iii) Calculate the acceleration of the cyclist when
he is instantaneously at rest.
(iv) Calculate the distance travelled by the cyclist
during the third second of his motion.

Ask by Daniels Martin. in Singapore

Dec 26,2024

Question

A cyclist travels in a straight line so that his
displacement, metres, from a fixed point , is
given by , where is the time in
seconds after passing a point on the line.
(i) Find the distance .
(ii) Find the velocity of the cyclist at .
(iii) Calculate the acceleration of the cyclist when
he is instantaneously at rest.
(iv) Calculate the distance travelled by the cyclist
during the third second of his motion.

Ask by Daniels Martin. in Singapore

Dec 26,2024

UpStudy AI · Accepted Answer

Let's solve each part of the problem step by step.

### Given:
Displacement from a fixed point $$ O $$:
$$ S(t) = 4 + 15t - t^3 $$
where $$ t $$ is the time in seconds after passing point $$ X $$.

---

### (i) Find the distance $$ OX $$.

**Understanding the Problem:**
- At $$ t = 0 $$, the cyclist is at point $$ X $$.
- $$ S(t) $$ gives the displacement from point $$ O $$.

**Solution:**
- Substitute $$ t = 0 $$ into the displacement equation:
  $$
  S(0) = 4 + 15(0) - (0)^3 = 4 \text{ meters}
  $$
  
**Answer:**
$$ OX = 4 \text{ meters} $$

---

### (ii) Find the velocity of the cyclist at $$ t = 2 $$.

**Understanding the Problem:**
- Velocity is the first derivative of displacement with respect to time.

**Solution:**
1. Differentiate $$ S(t) $$ with respect to $$ t $$:
   $$
   v(t) = \frac{dS}{dt} = 15 - 3t^2
   $$
2. Evaluate $$ v(t) $$ at $$ t = 2 $$:
   $$
   v(2) = 15 - 3(2)^2 = 15 - 12 = 3 \text{ m/s}
   $$

**Answer:**
$$ v(2) = 3 \text{ m/s} $$

---

### (iii) Calculate the acceleration of the cyclist when he is instantaneously at rest.

**Understanding the Problem:**
- Acceleration is the derivative of velocity with respect to time.
- The cyclist is instantaneously at rest when $$ v(t) = 0 $$.

**Solution:**
1. Find when $$ v(t) = 0 $$:
   $$
   15 - 3t^2 = 0 \implies 3t^2 = 15 \implies t^2 = 5 \implies t = \sqrt{5} \text{ seconds}
   $$
2. Differentiate $$ v(t) $$ to find acceleration $$ a(t) $$:
   $$
   a(t) = \frac{dv}{dt} = -6t
   $$
3. Evaluate $$ a(t) $$ at $$ t = \sqrt{5} $$:
   $$
   a(\sqrt{5}) = -6\sqrt{5} \text{ m/s}^2
   $$

**Answer:**
$$ a = -6\sqrt{5} \text{ m/s}^2 $$

---

### (iv) Calculate the distance traveled by the cyclist during the third second of his motion.

**Understanding the Problem:**
- "Third second" refers to the time interval from $$ t = 2 $$ seconds to $$ t = 3 $$ seconds.
- Distance traveled accounts for the total path length, irrespective of direction.

**Solution:**
1. **Determine if the cyclist changes direction between $$ t = 2 $$ and $$ t = 3 $$:**
   - Velocity at $$ t = 2 $$:
     $$
     v(2) = 3 \text{ m/s} \, (\text{positive, moving away from } O)
     $$
   - Velocity at $$ t = 3 $$:
     $$
     v(3) = 15 - 3(3)^2 = 15 - 27 = -12 \text{ m/s} \, (\text{negative, moving towards } O)
     $$
   - Since velocity changes sign, the cyclist changes direction during this interval.

2. **Find the time when $$ v(t) = 0 $$ within $$ 2 < t < 3 $$:**
   $$
   t = \sqrt{5} \approx 2.236 \text{ seconds}
   $$

3. **Calculate displacements:**
   - Displacement at $$ t = 2 $$:
     $$
     S(2) = 4 + 15(2) - (2)^3 = 4 + 30 - 8 = 26 \text{ meters}
     $$
   - Displacement at $$ t = \sqrt{5} $$:
     $$
     S(\sqrt{5}) = 4 + 15\sqrt{5} - (\sqrt{5})^3 = 4 + 15\sqrt{5} - 5\sqrt{5} = 4 + 10\sqrt{5} \approx 4 + 22.3607 = 26.3607 \text{ meters}
     $$
   - Displacement at $$ t = 3 $$:
     $$
     S(3) = 4 + 15(3) - (3)^3 = 4 + 45 - 27 = 22 \text{ meters}
     $$

4. **Calculate the distances:**
   - From $$ t = 2 $$ to $$ t = \sqrt{5} $$:
     $$
     \Delta S_1 = S(\sqrt{5}) - S(2) = 26.3607 - 26 = 0.3607 \text{ meters}
     $$
   - From $$ t = \sqrt{5} $$ to $$ t = 3 $$:
     $$
     \Delta S_2 = S(3) - S(\sqrt{5}) = 22 - 26.3607 = -4.3607 \text{ meters}
     $$
     (The negative sign indicates direction, but distance is positive.)
   - Total distance:
     $$
     \text{Total Distance} = |\Delta S_1| + |\Delta S_2| = 0.3607 + 4.3607 = 4.7214 \text{ meters}
     $$
   - For exact values:
     $$
     \text{Total Distance} = (10\sqrt{5} - 22) + (10\sqrt{5} - 18) = 20\sqrt{5} - 40 \approx 4.72 \text{ meters}
     $$

**Answer:**
$$ \text{Distance during the third second} \approx 4.72 \text{ meters} $$

---

### Summary of Answers:
1. **(i)** $$ OX = 4 $$ meters.
2. **(ii)** Velocity at $$ t = 2 $$ is $$ 3 $$ m/s.
3. **(iii)** Acceleration when instantaneously at rest is $$ -6\sqrt{5} $$ m/s².
4. **(iv)** Distance traveled during the third second is approximately $$ 4.72 $$ meters.
- **(i)** The distance from point $$ O $$ to point $$ X $$ is 4 meters. - **(ii)** The cyclist's velocity at 2 seconds is 3 m/s. - **(iii)** When the cyclist is momentarily at rest, their acceleration is $$ -6\sqrt{5} $$ m/s². - **(iv)** During the third second, the cyclist travels approximately 4.72 meters.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Calculus Questions