Question

A body covers 200 cm in the first 2 seconds and 220 cm in the next 4 seconds. Assuming constant acceleration, what is the velocity of the body at the end of the 7th second?
A) 40 cm/sec
B) 20 cm/sec
C) 10 cm/sec
D) 5 cm/sec

Answer 1

In this question, we need to determine the velocity of the body at the end of the 7th second. For this, we need to follow Newton's equations of motion. Moreover, here the journey of the body is divided into two parts so that two equations will be formed.

Complete step by step answer:
Following the newton’s second equation of motion $s = ut + \dfrac{{a{t^2}}}{2}$ where ‘s’ is the displacement of the body, ‘u’ is the initial velocity of the body, ‘a’ is the acceleration of the body, and ‘t’ is the instantaneous time.

Here, the traveling displacement of the body is divided into two parts:

Case 1: ${s_1} = 200{\text{ cm and }}{t_1} = 2{\text{ seconds}}$

So, substitute ${s_1} = 200{\text{ cm and }}{t_1} = 2{\text{ seconds}}$ in the formula $s = ut + \dfrac{{a{t^2}}}{2}$ to determine the relation between acceleration (a) and the initial velocity (u).

$s = ut + \dfrac{{a{t^2}}}{2} \\\ \Rightarrow 200 = u(2) + \dfrac{{a{{(2)}^2}}}{2} \\\ \Rightarrow 2u + 2a = 200 \\\ \Rightarrow u + a = 100 - - - - (i) \\\$

Case 2: ${s_2} = \left( {200 + 220} \right) = 420{\text{ cm and }}{t_2} = \left( {2 + 4} \right) = 6{\text{ seconds}}$

So, substitute ${s_2} = 420{\text{ cm and }}{t_2} = 6{\text{ seconds}}$ in the formula $s = ut + \dfrac{{a{t^2}}}{2}$ to determine the relation between acceleration (a) and the initial velocity (u).

$s = ut + \dfrac{{a{t^2}}}{2} \\\ \Rightarrow 420 = u(6) + \dfrac{{a{{(6)}^2}}}{2} \\\ \Rightarrow 6u + 18a = 420 \\\ \Rightarrow u + 3a = 70 - - - - (ii) \\\$

Now, solving the equation (i) and (ii) to determine the value of acceleration and the initial velocity.

From equation (i) we get,
$u + a = 100 \\\ \Rightarrow u = 100 - a - - - - (iii) \\\$

Substitute the expression for the initial velocity from the equation (iii) in the equation (ii) as:

$u + 3a = 70 \\\ \Rightarrow \left( {100 - a} \right) + 3a = 70 \\\ \Rightarrow 100 + 2a = 70 \\\ \Rightarrow 2a = 70 - 100 \\\ \Rightarrow a = \dfrac{{ - 30}}{2} \\\ = 15{\text{ cm/se}}{{\text{c}}^2} \\\$

Again, substitute the value of the acceleration in the equation (iii) to determine the value of the initial velocity as:

$u = 100 - a \\\ = 100 - ( - 15) \\\ = 115{\text{ cm/sec}} \\\$

Now, following Newton's first equation of motion $v = u + at$ where ‘v’ is the velocity of the body at the time ‘t’, ‘u’ is the initial velocity of the body, and ‘a’ is the acceleration of the body.

So, substitute $u = 115{\text{ cm/sec, }}a = - 15{\text{ cm/se}}{{\text{c}}^2}{\text{ and }}t = 7{\text{ sec}}$ in the formula $v = u + at$ to determine the velocity of the body at the end of the 7th seconds during its traveling period.

$v = u + at \\\ = 115 + ( - 15)7 \\\ = 115 - 105 \\\ = 10{\text{ cm/sec}} \\\$

Hence, the velocity of the body at the end of the 7th second is 10 cm/sec.

Option C is correct.

Note: It is very important to note here that it is given in the question that the acceleration while the body is moving is constant, and so, we have used the same acceleration during calculations. If the acceleration is varying then, we cannot use the same acceleration all over the calculation.