a) state whether the velocity of the car increasing at t=2 seconds, given the re

Question

a) state whether the velocity of the car increasing at t=2 seconds, given the reason for your answer.
b) At what time in the interval [0, 18], other than t=0, is the velocity of the car 60ft/sec? why?
c) On the time interval [0, 18], what is the car's absolute maximum velocity, in ft/sec, and at what time does it occurs? Justify your answer.
d) At what times in the interval 0< t <18 ( include 18), if any, is the car's velocity equal to 0? Justify your answer.

10. Below is a graph of the acceleration function a(t), in ft/sec2, of a car traveling on a straight road The ar initial velocity of 60 ft/sec. atr) (18, 15) (2. 15) is 2 468 10 12 14 A6 I (10.-13) (14.-15) 2 seconds, give the reason for your answer. (a) State whether the velocity of the car increasing at t act)a v(t)

Explanation / Answer

(a) Graph is representing acce;eration of car.

At t = 2 , accelerartion = 15 ft/sec^2

As we know acceleration is derivative of velocity.

So at t = 2 derivative of velocity is positive. In general if derivative of any function is positive then function is increasing.

So that's why velocity of car increasing at t = 2.

(b) From graph we can see at t = 6 and at t = 16 acceleration is zero.

So in general we know

v = u + at

where v = final velocity after t sec , u is intital velocity and a is acceleration.

When a = 0 then we can say

v = u . It means final velocity equal to initial velocity.

So in graph at t = 6 and tg = 16 acceleration is zero. So at these points velocity equal to its initial velocity 60 ft/sec.

(c) From graph at t = 6 and at t = 16 acceleration is zero means derivatve of velocity is zero.

So at these points maximum and minimum of function can occur.

At t = 6 slope of graph is negative means derivative of acceleration is negative. In other words we can say second derivative of velocity is negative. So we can conculde velocity is maximum at t= 6 because second derivative of velocity is negative at this point by second derivative test.

At t = 16 slope of aacceleration is positive. So second derivative of velocity is positive . So t = 116 relative minimum is occur.

(d) From graph we can conclude that when acceleration is positive or negative or when it is zero.

So from that we can conclude only when velocity is increasing or decreasing or remains constant.

SO we can not find where velocity is zero.

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