⇦ | ![]() | ⇨ |
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.
Related Questions:
- Length of a conductor is 50 cm radius of cross section 0.1 cm and resistivity
- Maximum velocity of the photoelectron emitted by a metal is 1.8×10⁶ ms⁻¹.
- When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid
- A bicyclist comes to a skidding stop in 10 m.During this process, the force
- To observed diffraction, the size of the obstacle
Difficulty Level: Easy
(1008)
Topics: Motion in Straight Line
(93)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply