A body having a constant acceleration travels200 cm in the first two seconds and 220 cm in thenext four seconds. What is its velocity at the endof 7 seconds.​

Answers (1)

A body having a constant acceleration travels200 cm in the first two seconds means

\begin{aligned} &\text { Here, } s=200 \mathrm{cm}, t=2 \mathrm{sec}\\ \end{aligned}

So using

 S=ut+\frac{1}{2}at^2\\\Rightarrow 200=2u+2a\\\Rightarrow u+a=100

Again the body covers  220 cm in the next four seconds

So total distance=s= 200+220=420 cm

total time =t=2+4=6s

So using

 S=ut+\frac{1}{2}at^2\\\Rightarrow 420=6u+18a\\\Rightarrow u+3a=70

From these 2 equations, we get

u=115 \mathrm{cm} / \mathrm{sec}, \text { and } a=-15 \mathrm{cm} / \mathrm{s}^{2}

Now  for t=7s

using

\begin{aligned} &\begin{array}{l} v=u+a t \\ =115-(15 \times 7) \end{array}\\ &=10 \mathrm{cm} / \mathrm{sec} \end{aligned}

 

 

Preparation Products

Knockout NEET May 2021 (One Month)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 14000/- ₹ 6999/-
Buy Now
Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Foundation 2021 Class 9th Maths

Master Maths with "Foundation course for class 9th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions