Read the following paragraph and answer the questions that follow:
Aditya started driving his car from rest. He increases the speed till 4 seconds and then he kept his car at
constant speed for 6 seconds. Then after that, he decreased the speed of the car up to another 6 seconds.
After 16 seconds of his driving, he applies the brakes to bring the car to rest.
(a) Plot the speed vs time graph for the 16 seconds journey of Aditya.
(b) Analyze the plotted graph and state what type of motion is performed by him for first 4 seconds of
the journey?
(c) What type of motion is represented by his motion from 10 seconds to 16 seconds?
(d) Find the acceleration of Aditya’s motion.
(e) Calculate the retardation of the journey.
(f) Find the distance travelled by Aditya for completing his journey.
plz any mod or brainly star answer
Answers
Answer:
a) Image of the graph is attached below.
b) The speed vs time graph for the situation is plotted as given in the image. It is given that Aditya starts from rest and keeps increasing his speed till 4 seconds. So this will be an accelerated motion with a positive slope.
c) The time from 10 to 16 is when he starts decreasing the speed of the car. So this will be decelerated motion.
For option d,e and f. Its actually not possible to find the value of acceleration and retardation from the given data. Maybe you should include some more details also to the question.
Let me try to give an generalized answer for d , e , f.
Let us assume he accelerates at a rate of a m/s² initially from rest.
After 4 seconds his speed will be:
v = 4a m/s (∵ v = u + at , u = 0 )
Then he moves for 6 seconds at constant velocity.
Now after 10 seconds of his total driving he starts decelerating for 6 seconds and comes to rest finally
⇒ 0 = 4a + 6a' (∵ v = 0 , a' - is the deceleration)
⇒ a' = - 2/3 a
Distance covered in part 1: (accelerated motion)
s = 0.5 at² = 8a m
Distance covered in part 2: (constant speed)
distance = speed × time
⇒ s₂ = 4a × 6 = 24a m
Distance covered in 3rd part: (decelerated motion)
s₃ = ut + 0.5 a't²
⇒ (4a)6 + 0.5 (-2/3 a) (6)²
⇒ s₃ = 24a - 12 a = 12a m
Total distance covered = 8a + 24a + 12a = 44a m .
Hope this helps : )
Answer:
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