Question

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.

Answer 1

k is 42.5 m/s

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

Answer 2

(a) 34 m/s

(b) k = 42.5

Explanation:

Part (a)

`\boxed{\sf Speed=(Distance)/(Time)}`

Given:

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

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

`\implies \sf Speed = (1700)/(50)=34\;m/s`

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.

`\boxed{\textsf{Area of a rectangle $=$ width $*$ length}}`

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

`\implies \sf 30k`

`\boxed{\textsf{Area of a triangle $=(1)/(2) *$ base $*$ height}}`

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

`\implies \sf (1)/(2)(20)k`

Therefore:

`\implies \sf 30k+(1)/(2)(20)k=1700`

`\implies \sf 30k+10k=1700`

`\implies \sf 40k=1700`

`\implies \sf (40k)/(40)=(1700)/(40)`

`\implies \sf k=42.5`