Question

A body travels 200 cm in the first two seconds and 220 cm in the next four seconds. What will be the velocity at the end of the 7th second from the start?

Answer 1

Hint: In this solution, we will use the relation of velocity, distance, and time in the equations of kinematics. We will first find the acceleration of the object and then use it to determine the velocity at the given time.
Formula used: In this solution, we will use the following formulae:
Second equation of kinematics: $d = ut + \dfrac{1}{2}a{t^2}$ where $d$ is the distance travelled by the object with initial velocity $u$ under acceleration $a$ in time $t$
First equation of kinematics: $v = u + at$ where $v$ is the final velocity of the object.

Complete step by step answer:
We’ve been told that a body travels 200 cm in the first two seconds and 220 cm in the next four seconds. Using the second law of kinematics for case 1,
$200 = 2u + 2a$
Which gives us,
$u + a = 100$
Similarly, for the second case, we will apply the second law of kinematics. In this case, the body has covered a total distance of $220 + 200 = 420\,cm$ in a total time of 6s. So, we can write
$420 = 6u + 18a$
$ \Rightarrow 70 = u + 3a$
Solving equation (1) and (2), we get
$2a = - 30\,cm/{s^2}$
And
$a = - 15\,m/{s^2}$ is the acceleration of the body. The initial velocity can be determined by substituting the value of acceleration in equation (1) as
$u - 15 = 100$
Which gives us
$u = 115\,m/s$
Then the velocity of the object at the 7th second can be determined by the first equation of kinematics as
$v = u + at$
Substituting $u = 115\,m/s$ and $t = 7$,we get
$v = 115 + ( - 15)(7)$

$v = 10\,m/s$ will be the velocity of the object at the 7th second.

Note: We must be careful that the distance travelled by the object is given for the first two seconds and then in the next four seconds. Since the acceleration is negative, the object is decelerating but since it has a non-zero initial velocity it will travel a finite distance.