A car accelerates from rest at constant rate for first 10s and covers a distance x.

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A car accelerates from rest at constant rate for first 10s and covers a distance x.  It covers a distance y in next 10s at the same acceleration.Which of the following is true?

Options

(a) x=3y
(b) y=3x
(c) x=y
(d) y=2x

Correct Answer:

y=3x

Explanation:

From equation of motion, we have S = ut+(1/2) at²
where, u=initial velocity, t=time, a=acceleration.
Since, car accelerates from rest u=0, t=10s
.·. s = 0 + (1/2) x a x (10)² ( .·. S=x)
x = (1/2) x a x (10)²
then, a = 2x/(10)²
Also
v = u + at Where, v is the final velocity.
= 0 + [2x/(10)²] x 10
= 2x/10ms
In the next 10s car moves with constant acceleration and with initial
velocity u = 2x/10 m/s
.·. S = (2x/10) x 10 + (1/2) x [2x/(10)²] x (10)²
= 2x + x when S=y
y = 3x
x and y is the two different distance covers a car with same time.

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Difficulty Level: Easy (1008)
Topics: Motion in Straight Line (93)
Subject: Physics (2479)

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