Question

A particle is moving along a straight line such that its velocity is increasing at $5\text{ ms}^{-1}$ per meter. When its velocity becomes $20\text{ ms}^{-1}$ its acceleration is:

• A. $50\text{ ms}^{-2}$
• B. $75\text{ ms}^{-2}$
• C. $100\text{ ms}^{-2}$
• D. $10\text{ ms}^{-2}$

Hint:

Remember that acceleration has two primary mathematical expressions:
$a = \frac{dv}{dt}$ (with respect to time) and
$a = v\frac{dv}{dx}$ (with respect to position).
Use the spatial derivative form when the rate is given "per meter".

Answer 1

The Correct Option is C

Step 1: Set up the chain rule.
We are told the velocity changes at a fixed rate with position, $\frac{dv}{dx} = 5\text{ s}^{-1}$. Acceleration is the rate of change of velocity with time, $a = \frac{dv}{dt}$.
Step 2: Rewrite using the chain rule.
Since $x$ is changing with time too, we can write $\frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = \frac{dv}{dx}\cdot v$. This builds the acceleration up from first principles instead of recalling it as a ready-made formula.
Step 3: Plug in the numbers.
At the instant $v = 20\text{ ms}^{-1}$, \[ a = 5\text{ s}^{-1} \times 20\text{ ms}^{-1} = 100\text{ ms}^{-2} \]
\[ \boxed{a = 100\text{ ms}^{-2}} \]