Question #104856

At a time t seconds after it is thrown up in the air, a tomato is

at a height (in meters) of f(t) = =4.9t

2 +50t +5 m.

A. What is the average velocity of the tomato during the

first 4 seconds? (Include units.)

B. Find (exactly) the instantaneous velocity of the tomato

at t = 4. (Include units.)

C. What is the acceleration at t = 4? (Include units.)

D. How high does the tomato go? (Include units.)

E. How long is the tomato in the air?

Expert's answer

f(t)=−4.9t2+50t+5f(t) = -4.9t ^2 +50t +5

f(0)=5f(4)=−4.9∗42+50∗4+5=126.6f(0)=5\\f(4)=-4.9*4^2+50*4+5=126.6


A)average velocity = f(4)−f(0)4−0\dfrac{f(4)-f(0)}{4-0} =121.6/4=30.4 m/s


B) instantaneous velocity = f′(4)f'(4) =−2∗4.9∗t+50=−2∗4.9∗4+50=10.8m/s=-2*4.9*t+50=-2*4.9*4+50=10.8 m/s


C) Acceleration =f′′(t)=d(f′(t))/dt=−2∗4.9f''(t)=d(f'(t))/dt=-2*4.9

f′′(4)=−2∗4.9=−9.8m/s2f''(4) =-2*4.9 =-9.8 m/s^2


D) Maximum height achieved by the tomato = maximum value of f(t)maximum \ value \ of \ f(t)

f′(t)=0−2∗4.9∗t+50=0t=25/4.9 sec.t=5.1 secf(5.1)=−4.9∗(5.1)2+50∗5.1+5=132.5mf'(t)=0\\ -2*4.9*t+50=0\\t=25/4.9 \ sec.\\t=5.1 \ sec \\f(5.1) =-4.9*(5.1)^2+50*5.1+5=132.5 m


E)When an object is thrown vertically upwards then time taken by the object to achieve the maximum height is equal to time taken by the object to return to its initial position from the maximum height

Time the tomato is in air is twice the time taken by the tomato to achieve the maximum height

Time = 2*5.1 =10.2 sec.


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