You observe an ant crawling up a sunflower stem. You estimate that its speed is

Question

You observe an ant crawling up a sunflower stem. You estimate that its speed is a constant 1.0cm/s and that is started crawling at this speed about 2.0cm above where the stem meets the ground.

A. Write down the ant's final position for some unknown time in the vertical direction. Take the ground to be the origin and assume it started crawling at the initial time of 0.

B. What's the ants position at t final= 6 seconds?

C. What was the ant's displacement between t final = 1 seconds and t final = 3 seconds

D. Now suppose the ant suddenly slips off the stem at t final = 6 seconds. Write a new kinematic equation for the ant's position, again using the ground as the origin.

E. When does the ant hit the ground?

F. What is its instantaenous speed just before it lands?

Explanation / Answer

Given

Speed of ant is constant, hence accelaration is zero

Constant speed, V = 1 cm/s

Ant started crawling at this speed at about 2 cm above the ground

A)

distance travelled by ant , s = v * t

or s = 1 * t

taking ground as origin , position of ant at any time t is given by

s = t + 2 cm

B)

at t = 6 s

s = 6 + 2 = 8 cm above the ground

C )

position of ant at t = 1 sec

s1 = 1 + 2 = 3 cm above the ground

position of ant at t = 3 sec

s2 = 3 + 2 = 5 cm above the ground

thus displacement s2 - s1 = 5-3

or displacement s2 - s1 = 2 cm

D)

now ant slips off at t= 6 sec

at t= 6sec , ant's position is s = 6+2 = 8cm above the ground

or s = 0.08 m

now

initial velocity for ant falling, v0 = 0

Since, S= v0t + 1/2*a*t2

or S = 0* t + 1/2*g*t2

or S = 1/2 * 9.8 * t2

or S = 4.9*t2

now new kinematic equation for position of the Ant is

position r = 0.08 - 4.9*t2   with ground as the origin

E)

Distance ant is above the ground, s = 0.08 m

let t be the time taken by the ant to hit the ground

S = 1/2*g*t2   since initial velocity is zero

0.08 = .5 * 9.8 * t2

t2 = 0.08 / 4.9 = 0.01632

or t = 0.127 sec

F)

Instantaneous speed can be measured by conservation of energy

so

let the velocity of ant just before hitting the ground be v

height above the ground when ant starts to fall, h = 0.08 m

now,

mgh = 1/2*m*v2

g*h = .5*v2

v2 = (9.8 * 0.08) / .5

v2 = 1.568

or v = 1.25 m/s

hence, instantaneous speed of the ant just before it hits the ground is 1.25 m/s

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.