Next, consider the case where 5t10 s, which is on the downward sloping portion o

Question

Next, consider the case where 5t10 s, which is on the downward sloping portion of the acceleration plot. The case is depicted in Interactive Fig. 2.14.4 below. The truck’s velocity change is the sum of the areas of the blue triangle, the tan rectangle, and the green triangle. We can use this graph to derive an expression for v(t) for 5t10 s.

Interactive Figure 2.14.4: An arbitrary instant t>5 s is shown. The total of the areas of the blue triangle, the tan rectangle, and the green triangle gives the velocity change for the interval from t=0 to t

The area of the large blue triangle is 12(5 s)(0.5 m/s2)=1.25 m/s . The area of the tan rectangle is (10.1t)(t5) . The area of the green triangle is 12(t5)[0.5(10.1t)] . The sum of these areas is the total velocity change for the interval from t=0 to t where 5t10 s :

0t adt=(1.25)+(10.1t)(t5)+12(t5)[0.5(10.1t)]

0t adt=(1.25)+(10.1t)(t5)+12(t5)(0.5+0.1t)

0t adt=(1.25)+(t50.1t2+0.5t)+12(0.5t+0.1t2+2.50.5t)

0t adt=(1.25)+(5+1.5t0.1t2)+12(2.5t+0.1t2)

0t adt=2.5+t0.05t2

The velocity of the truck at the instant t follows from v=v1+0tadt , with proper SI units included:

v=v1+0tadt

v=(1 m/s)+(2.5 m/s)+(1.0 m/s2)t(0.05 m/s3)t2

Since the truck’s initial velocity is a negative number, and the acceleration is positive for 0<t<10 s, you should expect the truck to slow down initially, stop for an instant, and then speed up while moving in the +x direction. This is exactly what the velocity versus time plot tells us happens!

At what instant will the truck have its most negative position? t=?s

Interactive Figure 2.14.4: An arbitrary instant t>5 s is shown. The total of the areas of the blue triangle, the tan rectangle, and the green triangle gives the velocity change for the interval from t=0 to t

The area of the large blue triangle is 12(5 s)(0.5 m/s2)=1.25 m/s . The area of the tan rectangle is (10.1t)(t5) . The area of the green triangle is 12(t5)[0.5(10.1t)] . The sum of these areas is the total velocity change for the interval from t=0 to t where 5t10 s :

0t adt=(1.25)+(10.1t)(t5)+12(t5)[0.5(10.1t)]

0t adt=(1.25)+(10.1t)(t5)+12(t5)(0.5+0.1t)

0t adt=(1.25)+(t50.1t2+0.5t)+12(0.5t+0.1t2+2.50.5t)

0t adt=(1.25)+(5+1.5t0.1t2)+12(2.5t+0.1t2)

0t adt=2.5+t0.05t2

The velocity of the truck at the instant t follows from v=v1+0tadt , with proper SI units included:

v=v1+0tadt

v=(1 m/s)+(2.5 m/s)+(1.0 m/s2)t(0.05 m/s3)t2

v=3.5 m/s+(1.0 m/s2)t(0.05 m/s3)t2  , 5 st10 s (A) What is this area? equation of this line segment: 1-0.1t equation of this line segment: 0.1 a (m/s2) 0.5 0.4 0.3 0.2 o.h 10 What is this area What is this area

Explanation / Answer

it can be answered in two ways.

First, the x wil go on to be negative as long as v i s negative. i.e v is less than zero.
From graph, v = 0 at t = 4.5s

Second, x will be min when dx/dt = 0

but dx/dt = v

0 = -1 -2.5 + 1 t - 0.05 t^2

t = 4.523 sec

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