Question

Each of the four particles move along $x$ axis. Their coordinate (in meters) as function of time (in seconds) are given by
Particle 1: $x\left(t\right)=3.5-2.7t^3$
Particle 2: $x\left(t\right)=3.5+2.7t^3$
Particle 3: $x\left(t\right)=3.5+2.7t^2$
Particle 4: $x\left(t\right)=3.5-3.4t-2.7t^2$
Which of these particles are speeding up for $t>0$?

• A. All four
• B. Only 1
• C. Only 2 and 3
• D. Only 2, 3 and 4

Answer 1

The correct option is A All four

At $t>0$
For particles 2 and 3
$\frac{dx}{dt}>0$
Also, $\left|\frac{d^{2}x}{d{t}^2}\right|>0$
As, both velocity and acceleration are positive so particle will be speeding up
For particle 1
At $t>0$
$\frac{dx}{dt}=-8.1t^2$
$\Rightarrow \frac{dx}{dt}<0$
and $\frac{d^{2}x}{d{t}^2}=-16.2t$
$\therefore \frac{d^{2}x}{d{t}^2}<0$
As, both velocity and acceleration are negative so in this case also body will speed up
For particle 4
At $t>0$
$\frac{dx}{dt}=-3.4-5.4t$
$\Rightarrow \frac{dx}{dt}<0$
and $\frac{d^{2}x}{d{t}^2}=-5.4$
$\therefore \frac{d^{2}x}{d{t}^2}<0$
As, both velocity and acceleration are negative so in this case also body will speed up
Therefore for $t>0;\left|\frac{dx}{dt}\right|$ is increasing in all.