A chameleon extends its tongue to capture a tasty insect. The tongue accelerates at 33 m/s2 for 0.12 seconds to make the capture. what is the speed of the chameleons tongue when it grabs the insect?

so the answer is v= 3.96?

1/4 mile = 402 meters

average speed = (final speed + original speed)/2 if acceleration is constant

v = 140.3/2
= 70.15 m/s

time = distance/speed = 402/70.15 = 5.73 seconds

check
v end = 0 + 30.4 (5.73)?
174 not really

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car 1 a = 10/10 = 1
car 2 a = 0/15 = 2
car 3 a = -20/5 = -4

so 3 , 1 , 2

v = a t

= 33 (.12)

the winner of the drag race was traveling with a speed of 140.3 m/s (313.9 mph) at the end of the quarter mile course the car started from rest and had an average acceleration of 30.4 m/s^2. what was the winning time for this race

v=140.3 m/s
a=30.4 m/s^2
t=?

140.4/30.4= 4.6 m/s

i don't understand how u did this??

To find the speed of the chameleon's tongue when it grabs the insect, we can use the equation of motion:

v = u + at

Where:
v = final velocity (speed)
u = initial velocity (which is usually zero in these types of problems)
a = acceleration
t = time

In this case, the initial velocity (u) is zero because the tongue starts from rest. The acceleration (a) is given as 33 m/s^2, and the time (t) is 0.12 seconds.

Substituting these values into the equation, we have:

v = 0 + (33 m/s^2)(0.12 s)
v = 0 + 3.96 m/s
v ≈ 3.96 m/s

Therefore, the speed of the chameleon's tongue when it grabs the insect is approximately 3.96 m/s.

three cars accelerate while moving in the positive direction rank the cars in order of increasing acceleration from most negative to most positive.

*car 1 speeds up from 25 m/s to 35 m/s in 10 sec
* car 2 speeds up from 0 to 30 m/s in 15 sec
*car 3 slows down from 32 m/s to 12 m/s in 5 sec

answer 3,2,1???