Question

A car accelerates from rest at constant rate for first 10 s and covers a distance x.It covers a distance y in next 10 s at the same accleration.Which of the following is true?
(a) x=3y (b) y=3x (c) x=y (d) y=2x

Answer 1

Explanation:

Intial Speed=0

Now Car move ar Constant Rate for 10 s

and cover distance x.

It cover y in next 10s at the Same Accerlation.

Now

Final Velocity in First case

v=u+at

v=at

v=10a

S=ut+1/2at²

S=0+1/2at²

S=1/2×a(10)²

S=50a (1)

Distance Travelled in First case=50a

Now 10a is Intial Speed for Second case

Accerlation=a

T=10s

Distance=y

v=at=10a (2)

t=10a/a

t=10. (3)

So,

Distance in Second Case.

S=ut+1/2at²

y=10a(10)+1/2×a(10)²

y=100a+50a

y=150a (4)

Dividing Equation (4) from Equation (1)

Y=150a

X=50a

Y/X=150a/50a

Y/X=3

Y=3X

Relation 2 is right

Y=3X

Answer 2

Initial speed (u) = 0

According to question

1st case

v = u + at

v = 0 + a(10)

v = 10a

Now

Distance = x

x = ut + 1/2at²

x = 0 + 1/2a(10)²

x = 50a

Initial Speed for 2nd case = 10a

Acceleration = a

t = 10s

Distance = y

v = u + at

v = 10a + 10a

v = 20a

Distance in Second Case.

s = ut + 1/2at²

y = 10a(10) + 1/2a(10)²

y = 100a + 50a

y = 150a

Now;

y/x = 150a/50a

y/x = 3

y = 3x