Small frogs that are good jumpers are capable of remarkable accelerations. one species reaches a takeoff speed of3.7 m/s in60 ms . part a what is the frog's acceleration during the jump?

We know that acceleration is change in velocity by time taken for that change.

In this case velocity change is 3.7 m/s

Time taken for this change = 60 ms =

So acceleration of frog  =

                                       = 61.66 m/

So acceleration of frog is 61.66 m/

o it is evident that frog is capable of remarkable accelerations.

The frog's acceleration during the jump is 62 m/s²

Acceleration is rate of change of velocity.

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem !

Given:

initial speed = u = 0 m/s

final speed = v = 3.7 m/s

time taken = t = 60 ms = 0.06 s

Unknown:

acceleration = a = ?

Solution:

To find the acceleration of th frog, it can be done in the following way:

The frog's acceleration during the jump is 62 m/s²

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate, Frog , Jump

c. 7/3

step-by-step explanation:

a rational number is like pi- it is a repeating decimal.

a. is -1.8

b. 2.236068

c. 2.3333333333333333333

d.   is 3

i hope that ! =)

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length