Question

The equation $\mathrm{s}=3{\mathrm{t}}^3+2{\mathrm{t}}^2+\mathrm{t}+1$ gives the distance (in meters) traveled by a vehicle in time $\mathrm{t}$ (in seconds). The acceleration difference between$\mathrm{t}=2$ and $\mathrm{t}=4$is

• A. $36\mathrm{m}/\mathrm{s}2$
• B. $38\mathrm{m}/\mathrm{s}2$
• C. $45\mathrm{m}/\mathrm{s}2$
• D. $46\mathrm{m}/\mathrm{s}2$

Answer 1

The correct option is A

$36\mathrm{m}/\mathrm{s}2$

Step: 1 Given data:

$\mathrm{s}=3{\mathrm{t}}^3+2{\mathrm{t}}^2+\mathrm{t}+1$

Step: 2 Formula used:

The velocity of the particle $\left(\mathrm{v}\right)=\frac{\mathrm{ds}}{\mathrm{dt}}$

The acceleration of the particle $\left(\mathrm{a}\right)=\frac{\mathrm{dv}}{\mathrm{dt}}$

Step: 3 Calculate the acceleration at $\mathrm{t}=2$and $\mathrm{t}=4$:

Velocity $\left(\mathrm{v}\right)=\frac{\mathrm{ds}}{\mathrm{dt}}=\frac{\mathrm{d}\left(3{\mathrm{t}}^3+2{\mathrm{t}}^2+\mathrm{t}+1\right)}{\mathrm{dt}}=9{\mathrm{t}}^2+4\mathrm{t}+1$

Acceleration $\left(\mathrm{a}\right)=\frac{\mathrm{d}\left(9{\mathrm{t}}^2+4\mathrm{t}+1\right)}{\mathrm{dt}}=18\mathrm{t}+4$

Acceleration at $\mathrm{t}=2$

${\left(\mathrm{a}\right)}_{\mathrm{t}=2}=18\left(2\right)+4=40$

Acceleration at $\mathrm{t}=4$

${\left(\mathrm{a}\right)}_{\mathrm{t}=4}=18\left(4\right)+4=76$

Difference between $\mathrm{t}=2$ and $\mathrm{t}=4$

$$\begin{gathered} \mathrm{a}={\left(\mathrm{a}\right)}_4-{\left(\mathrm{a}\right)}_2 \\ \mathrm{a}=76-40 \\ =36\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, option A is correct.