Please, i need help for part b, my hw is due tomorrow, please help, any help will be appreciated,

The graph shows the speed of a vehicle during the final 50 seconds of a journey.
At the start of the 50 seconds the speed is k metres per second.
The distance travelled during the 50 seconds is 1.7 kilometres.


b) Work out the value of k.

Please i need help for part b my hw is due tomorrow please help any help will be appreciated The graph shows the speed of a vehicle during the final 50 seconds class=


Answer :

Answer:

(a)  34 m/s

(b)  k = 42.5

Step-by-step explanation:

Part (a)

[tex]\boxed{\sf Speed=\dfrac{Distance}{Time}}[/tex]

Given:

  • Distance = 1.7 km = 1700 m
  • Time = 50 s

Substitute the values into the formula to find the average speed :

[tex]\implies \sf Speed = \dfrac{1700}{50}=34\;m/s[/tex]

Part (b)

The area under a speed-time graph represents the distance traveled.

Separate the area under the graph into a rectangle and a triangle, where:

  • The area of the rectangle represents the distance traveled in the first 30 seconds of the journey.
  • The area of the triangle represents the distance traveled in the last 20 seconds of the journey.

[tex]\boxed{\textsf{Area of a rectangle $=$ width $\times$ length}}[/tex]

Therefore, the distance traveled in the first 30 seconds of the journey is:

[tex]\implies \sf 30k[/tex]

[tex]\boxed{\textsf{Area of a triangle $=\dfrac{1}{2} \times$ base $\times$ height}}[/tex]

Therefore, the distance traveled in the last 20 seconds of the journey is:

[tex]\implies \sf \dfrac{1}{2}(20)k[/tex]

Therefore:

[tex]\implies \sf 30k+\dfrac{1}{2}(20)k=1700[/tex]

[tex]\implies \sf 30k+10k=1700[/tex]

[tex]\implies \sf 40k=1700[/tex]

[tex]\implies \sf \dfrac{40k}{40}=\dfrac{1700}{40}[/tex]

[tex]\implies \sf k=42.5[/tex]

Answer:

k is 42.5 m/s

Step-by-step explanation:

we need to calculate the distance that was covered when the speed was k m/s

from the graph k m/s was travelled for 30 seconds

the entire time for the journey was 50 seconds

the entire journey was 1700 m

Alternatively the area under the graph represents the total distance covered.

Area of a trapezium = 1/2(a+b)h

= 1/2( 30+50)k

= 40k

we equate it to the total distance covered

1700 = 40k.

k = 42.5

Thus k is 42.5 m/s