mMCQ.

Navigation Menu

Step into mMCQ.

Launch mMCQ. diagnostic

Explore mMCQ.

MDCAT prepFree DiagnosticPricing & SubscribeSign in

Resources

Terms & Conditions

mMCQ.

© 2021 - 2025 mMCQ.All rights reserved.

WhatsApp
  1. Home
  2. MDCAT
  3. Physics
  4. Forces and Motion
MDCAT
Physics
2017

Physics · Forces and Motion

Work through this past-paper style MCQ, then read the full explanation. Practice more physics questions on mMCQ with adaptive practice and topic analytics.

Question

The distance travelled by a moving car with velocity 15m/s in 2s, decelerates at 2m/s2 is equal to:

Options
  • A

    30m

  • B

    16m

  • C

    34m

  • D

    26m

Explanation

This is a question from a body moving at constant acceleration for that we have 3 equations of motion 

vf=vi + at

S=vit +1/2at2

2aS=vf2 – vi2

Here S refers to a displacement

Vi is the initial velocity 

Vf is the final velocity 

t is for time 

a is for acceleration 

According to the data given, the most suitable equation to be used here is the 2nd equation of motion,

SOL: S=vit+1/2at2

            =15x2+½x(-2)22

-ve sign indicates that the car is decelerating 

            =30–4

            =26m

Hence the correct answer is option D.

As per the above calculation, this option is incorrect.

According to the data given, the most suitable equation to be used here is the 2nd equation of motion,

SOL: S=vit+1/2at2

=15x2+½x(-2)22

-ve sign indicates that the car is decelerating

=30–4

=26m

As per the above calculation, this option is incorrect.

According to the data given, the most suitable equation to be used here is the 2nd equation of motion,

SOL: S=vit+1/2at2

=15x2+½x(-2)22

-ve sign indicates that the car is decelerating

=30–4

=26m

As per the above calculation, this option is incorrect.

According to the data given, the most suitable equation to be used here is the 2nd equation of motion,

SOL: S=vit+1/2at2

=15x2+½x(-2)22

-ve sign indicates that the car is decelerating

=30–4

=26m

This is a question from a body moving at constant acceleration for that we have 3 equations of motion 

vf=vi + at

S=vit +1/2at2

2aS=vf2 – vi2

Here S refers to a displacement

Vi is the initial velocity 

Vf is the final velocity 

t is for time 

a is for acceleration 

According to the data given, the most suitable equation to be used here is the 2nd equation of motion,

SOL: S=vit+1/2at2

            =15x2+½x(-2)22

-ve sign indicates that the car is decelerating 

            =30–4

            =26m

Hence the correct answer is option D

Keep practising Physics

Take the free 2 minutes diagnostic to map weak topics, then unlock full Biology, Chemistry, Physics, English, and Logical Reasoning practice for PKR 15,000/month.

Start free diagnosticExplore MDCAT

Tagged under Physics · Forces and Motion · 2017