A particle is moving along the parabola y x2 8 = When the particle is at (3, 9/8) and is moving such that its x-coordinate is decreasing, it is observed that its speed is 5 m/s and is increasing at a rate of 2.4 m/s². Find the acceleration of the particle at this instant.a. 1b. 2c. 3d. 4

1. Map the Track

The path is fixed by

$y = \frac{x^2}{8}$

Hence position vector

$\vec r = x\,\hat i + \frac{x^2}{8}\,\hat j$

2. Link $x$ to Velocity

Differentiate with respect to time $t$:

$\vec v = \frac{dx}{dt}\,\hat i + \frac{x}{4}\,\frac{dx}{dt}\,\hat j = x'\!\left(\hat i + \frac{x}{4}\,\hat j\right)$

where $x' = \dfrac{dx}{dt}$.
Speed (magnitude)

$v = |\vec v| = |x'|\sqrt{1 + \left(\frac{x}{4}\right)^2}$

3. Plug the Instantaneous Data

At the point $(x,y)=(3,9/8)$:

$\sqrt{1 + \left(\frac{3}{4}\right)^2}=\sqrt{1+0.5625}=1.25$

Given speed $v=5$ m/s,

$|x'|=\frac{5}{1.25}=4 \quad\Longrightarrow\quad x'=-4\;\text{m/s (negative because $x$ is decreasing)}$

So

$\vec v = -4\,\hat i-3\,\hat j$

4. Find the General Acceleration Vector

Differentiate $\vec v$:

$\vec a = \frac{d^2x}{dt^2}\,\hat i + \left(\frac{x}{4}\,\frac{d^2x}{dt^2}+\frac{(x')^2}{4}\right)\hat j = x''\,\hat i + \left(\frac{x}{4}x''+\frac{(x')^2}{4}\right)\hat j$

where $x'' = \dfrac{d^2x}{dt^2}$.

5. Use the Given Tangential (speed-changing) Information

Tangential (along-velocity) component obeys

$\frac{dv}{dt}=\frac{\vec v\,\cdot\,\vec a}{v}$

Given $dv/dt = 2.4$ m/s².

Compute $\vec v\cdot\vec a$ at $x=3$:

• $a_x = x''$
• $a_y = \dfrac{3}{4}x'' + \dfrac{16}{4}=\dfrac{3}{4}x'' +4$

So

$\vec v\cdot\vec a = (-4)x'' + (-3)\left(\frac{3}{4}x''+4\right) = -\frac{25}{4}x'' -12$

Set up the tangential equation:

$\frac{-\tfrac{25}{4}x''-12}{5}=2.4\quad\Longrightarrow\quad -\frac{25}{4}x''-12=12$

$-\frac{25}{4}x''=24\quad\Longrightarrow\quad x''=-\frac{96}{25}\approx-3.84\;\text{m/s}^2$

6. Assemble the Acceleration Components

$a_x = -3.84$

$a_y = \frac{3}{4}(-3.84)+4 = -2.88+4 = 1.12$

7. Magnitude of Acceleration

$a = \sqrt{(-3.84)^2 + (1.12)^2}=\sqrt{14.7456+1.2544}=\sqrt{16}=4\;\text{m/s}^2$

Hence the required acceleration is 4 m/s² (option d).

Imagine this!

You have a shiny toy car that must stay on a curved road shaped exactly like a banana-shaped line called a parabola.
The rule of the road is

$y = \frac{x^2}{8}$

So wherever the car goes, its position $(x, y)$ must satisfy that little maths rule.

Now picture the car when it is exactly at the point $(3,\;9/8)$ on the road.

• It is driving “to the left” (so its $x$ value is getting smaller).
• Your speed-gun says its speed is 5 m/s.
• The car is pressing the accelerator so the speed is going up by 2.4 m/s every second.

Mummy asks, “Hey kid, what is the car’s total acceleration right now?”
(In easy words: How hard is it actually speeding up and curving?)

After working out the left–right and up–down pushes separately, you’ll discover the total push (acceleration) is 4 m/s².

Step-by-Step Solution

1. Path equation: $y=\dfrac{x^2}{8}$
2. Position vector: $\vec r = x\,\hat i + \frac{x^2}{8}\,\hat j$
3. Velocity: $\vec v = x'\,\hat i + \frac{x}{4}x'\,\hat j$
4. Speed relation: $v = |x'|\sqrt{1+\left(\frac{x}{4}\right)^2}$
   With $x=3$ and $v=5$: $|x'| = \frac{5}{1.25}=4\;\Rightarrow\;x'=-4$
5. Acceleration: $\vec a = x''\,\hat i + \left(\frac{x}{4}x'' + \frac{(x')^2}{4}\right)\hat j$
6. Dot product with velocity: $\vec v\cdot\vec a = (-4)x'' + (-3)\left(\frac{3}{4}x''+4\right) = -\frac{25}{4}x'' -12$
   Use $\dfrac{dv}{dt} = 2.4$: $\frac{-\tfrac{25}{4}x''-12}{5}=2.4\;\Rightarrow\;x''=-\frac{96}{25}\,(\approx -3.84)$
7. Components of $\vec a$: $a_x=-3.84,\quad a_y=\frac{3}{4}(-3.84)+4=1.12$
8. Magnitude: $a=\sqrt{(-3.84)^2+1.12^2}=4\;\text{m/s}^2$
   Answer: 4 (Option d)