A car start from rest and accelerates at the rate of 2m/s² for 5 seconds. it maintenance this velocity for another 5 seconds. it eventually undergoes a uniform deceleration and finally comes to rest after further 5 seconds. A) draw the velocity-time graph for the journey

B) find the acceleration and deceleration respectively,

C) find the total distance travelled during each stage.

D) find the average velocity over the entire period

2m/s² ; - 2m/s²

25m ; 50m ; 25m

6.67 m/s²

Step-by-step explanation:

The acceleration and deceleration :

Both rates are the Same from the graph ;

According to scenario described, acceleration = 2m/s²

Deceleration occurred in equal time interval = - 2m/s

Distance covered during each stage:

Stage 1 : area of triangle

0.5 * base * height

0.5 * 5 * 10 =25 m

Stage 2: area of rectangle :

Length * breadth = 5 * 10 = 50 m

Stage 3 : area of triangle

0.5 * base * height

0.5 * 5 * 10 =25 m

Average Velocity :

Total distance / time = (25 + 50 + 25) / 15 = 100/ 15 = 6. 67 m/s

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

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step-by-step explanation:

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