Question

The x and y coordinates of a particle at any time t are given by

x = 7t + 4t² and y = 5t,

Where x and y are in metres and t in seconds. The acceleration of a particle at t = 5s is.

• A. Zero
• B. 8 m/s²
• C. 20 m/s²
• D. 40 m/s²

Answer 1

The correct option is B

8 m/s²

Step 1: Given:

x and y coordinates of a particle at any time t are

x = 7t + 4t² and y = 5t

Where x and y are in metres and t in seconds.

Let t = 5 seconds.

Step 2: Calculating the total acceleration of the particle:

$\mathrm{total}\mathrm{acceleration}=\mathrm{a}=\sqrt{{{\mathrm{a}}_{\mathrm{x}}}^2+{{\mathrm{a}}_{\mathrm{y}}}^2}$

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}=\frac{\mathrm{d}\left(7\mathrm{t}+8{\mathrm{t}}^2\right)}{\mathrm{dt}}=7+8\mathrm{t}\mathrm{m}/\mathrm{s} \\ \mathrm{and},{\mathrm{a}}_{\mathrm{x}}=\frac{{\mathrm{dv}}_{\mathrm{x}}}{\mathrm{dt}}=\frac{\mathrm{d}\left(7+8\mathrm{t}\right)}{\mathrm{dt}}=8\mathrm{m}/{\mathrm{s}}^2........(\mathrm{independent}\mathrm{of}\mathrm{time}) \end{gathered}$$

$$\begin{gathered} \mathrm{Similiarly},{\mathrm{a}}_{\mathrm{y}}=\frac{{\mathrm{d}}^2\left(5\mathrm{t}\right)}{\mathrm{dt}}=0\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Hence}, \\ \mathrm{a}=8\mathrm{m}/{\mathrm{s}}^2. \end{gathered}$$

Hence, option B is the correct answer