Question

20 A graph for v - tof an object is shown below 5 4.6 А. B 4 3 Velocity (m/s) 2 Vo + 0 1 2 3 4 5 6 7 8 9 10 11 12 Time (s) (0) (ii) Identify the type of motion by line OA and BC. What is the time duration during which the car moves with constant velocity? Calculate the acceleration between the 3rd and 10th second. (iii) (iv) What is the distance travelled in the first 10 s?


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Answer 1

Answer:

The distance travelled in the first 10sec is 650m.

Hope it's helpful☺ ❤

Answer 2

Explanation:

i) For OA the type of motion is increasing constant velocity, and for BC is decreasing constant velocity

ii) we have to find the time of OA and BC

So that the time is 2 + 2 = 4s.

iii) The motion between AB is uniform velocity so there is 0 acceleration so in the time period of 3rd to 10th second the acceleration = 0.

iv) to find the distance we have to find the area under motion

for first triangle area = 1/2 × 2 × 4.6 = 4.6m.

for square area = 8 × 4.6= 36.8m.

for second triangle area = 1/2 × 2 × 4.6= 4.6m.

total area = 4.6 + 4.6 + 36.8 = 46m.

you can find area by other method also

eg. area of OABC = 1/2 × 4.6 × ( 12 + 8 ) = 46m.