Aditya started driving his car. He increased the speed till 4 seconds and then he kept his car in constant speed for 6 seconds.Then after he decreased the speed of the car up to another 6 seconds. After reaching at the starting place, he draws the speed- time graph of his 16 seconds driving as shown below:
Answers
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 .
