Such a jerk!

Classical Mechanics Level 3

Consider an object on which a non-constant force is applied. If it has initial velocity as $5 m/s$ , initial acceleration of $5 m/s^{2}$ & a jerk of $5 m/s^{3}$ . Find the distance the body travels in 12 seconds.

Details and assumptions
Jerk (J) is the rate of change of acceleration/the third derivative of displacement.

This problem is a part of my set The Best of Me

The answer is 1860.

2 solutions

Chew-Seong Cheong
Jul 29, 2014

It is given that $\dddot{s}=5$ $m/s^3$ , $\ddot{s}=5$ $m/s^2$ , and $\dot{s}=5$ $m/s$ , and we need to find $s$ , when time $t=12$ $s$ .

We do so by first integrating $\dddot{s}=5$ to get $\ddot{s}$ : $\ddot{s} = 5t + c$ As $\ddot{s}(0)=5$ , $c=5$ , and $\ddot{s} = 5t + 5$ Similarly, $\dot{s} = \frac{5}{2}t^2 + 5t + 5$ $s = \frac{5}{6}t^3+\frac{5}{2}t^2 + 5t$ Therefore, $s(12) = \frac{5}{6}\times 12^3+\frac{5}{2}\times 12^2 + 5\times 12$ $s(12) = 1440+360 + 60 = \boxed {1860}$

Yes.. @Siddharth G -Dont you think this is more a calculus problem that Mechanice?

Krishna Ar - 6 years, 9 months ago

Well, Mechanics was started as such by Sir Issac Newton. He and Leibniz invited calculus at the same time.

Chew-Seong Cheong - 5 years, 5 months ago

Yes exactly same way.

Kushagra Sahni - 5 years, 5 months ago

Prakhar Prakash
May 22, 2014

given that da/dt=5 . integrate get 'a' as function of t (remember while integrating put the correct limits). we get a=5t+5. now integrate this expression w.r.t. t and get v=5t^2/2 + 5t +5. again integrate to get the expression of position vs time. put the limits t=0 to t=12 and get s=1860.

Such a jerk!

The answer is 1860.

2 solutions

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