
A body travels 200 cm in the first 2 seconds and 220 cm in next 4 seconds. What will be the velocity at the end of the 7th second of the start?A. 2 cm/sB. 8 cm/sC. 10 cm/sD. 15 cm/s
Answer
588.9k+ views
Hint: First of all, find the initial velocity and acceleration of the body by using the second equation of motion. Then find the velocity at 7th sec by using the first equation of motion and the obtained values of initial velocity and the acceleration of the body.
Formulae used: \[s = ut + \dfrac{1}{2}a{t^2}\], \[v = u + at\]
Complete step-by-step solution -
Formulae used: \[s = ut + \dfrac{1}{2}a{t^2}\], \[v = u + at\]
Given at time \[t = 2{\text{ sec}}\], distance covered by the body \[s = 200{\text{ cm}}\]
Complete step-by-step solution -
We know the formulae for second equation of motion i.e., \[s = ut + \dfrac{1}{2}a{t^2}\] where \[s\] is the distance covered, \[u\] is the initial speed of the body, \[t\] is the time taken and \[a\] is the acceleration of the body. So, at time \[t = 2{\text{ sec}}\] and \[s = 200{\text{ cm}}\], we have
$200 = u\left( 2 \right) + \dfrac{1}{2}a{\left( 2 \right)^2}$
$\Rightarrow 200 = 2u + 2a $
$\Rightarrow 100 = u + a $
$\Rightarrow u = 100 - a........................................................................\left( 1 \right) $
Also, the body covers a distance of 220 cm in the next 4 sec.So, we have \[t = 2 + 4 = 6{\text{ sec}}\] and \[s = 200 + 220 = 420\], we have
$ 420 = u\left( 6 \right) + \dfrac{1}{2}a{\left( 6 \right)^2} $
$\Rightarrow 420 = 6u + 18a $
$\Rightarrow 70 = u + 3a $
$\Rightarrow u = 70 - 3a..............................................................\left( 2 \right)$
From equation (1) and (2), we have
$100 - a = 70 - 3a $
$\Rightarrow 3a - a = 70 - 100 $
$\Rightarrow 2a = - 30 $
$\Rightarrow a = - 15{\text{ cm/}}{{\text{s}}^2}$
By substituting \[a = - 15\] in equation (1) we get
$u = 100 - \left( { - 15} \right) = 100 + 15 $
$\Rightarrow u = 115{\text{ cm/s}} $
From the equations of motion, we know that \[v = u + at\] where \[v\] is the final velocity of the body, \[u\] is the initial velocity of the body, \[a\] is the acceleration of the body and \[t\] is the time taken by the body to cover a particular distance.
So, the velocity at the end of 7th second i.e., at \[t = 7{\text{ sec}}\], we have
$\Rightarrow v = 115 + \left( { - 15} \right)7 $
$\Rightarrow v = 115 - 105 $
$\therefore v = 10{\text{ cm/s}} $
Thus, the correct option is C. 10 cm/s.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

