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a car start from rest and accelerates at the rate of 2ms² 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 ​

Sagot :

fichoh

Answer:

Acceleration = 2m/s ; deceleration = - 2 m/s

Distance covered :

Stage 1 = 25

Stage 2 = 50 m

Stage 3 = 25m

Average Velocity = 6.67 m/s

Step-by-step explanation:

Given that:

Acceleration = 2m/s

Final velocity, v

v = u + at

u = 0 (from rest)

v = 0 + 2(5)

v = 10 m/s

deceleration ;

Final velocity, v ;

v = u + at

v = 0 ( to rest)

0 = 10 + a * 5

0 = 10 + 5a

-5a = 10

a = 10 / - 5

a = - 2 m/s^2

Distance traveled during each stage :

Distance = speed * time

Stage1 = S = 0.5at^2 = 0.5 * 2 * 5^2 = 0.5 * 2 * 25 = 25 meters

Stage 2 = Distance = speed * time = 10m/s * 5 = 50 metres

Stage 3 = distance = S = 0.5at^2 = 0.5 * 2 * 5^2 = 0.5 * 2 * 25 = 25 meters

Average Velocity :

Total distance traveled / total time taken

(25 + 50 + 25)m / 15s

100m / 15 s

Average Velocity = 6.667 m/s

View image fichoh
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