Question

A particle starts from rest with constant acceleration for 20 seconds. If it travels a distance y 1 in the first 10 seconds and a distance y 2 in the next 10 seconds then?

$$B){y_2} = 3{y_1} \\\ C){y_2} = 4{y_1} \\\ D){y_2} = 5{y_1} \\\$$

Answer 1

Hint: Here, the particle starts at rest, so its initial velocity is zero. The final velocity is not given, but the distance it travels for the first 10 seconds and the second 10 seconds at constant acceleration can be known through the second equation of linear motion of kinematics at constant acceleration. Then we take a ratio of the distances obtained.

Complete step by step solution:
Here, let the total distance covered by the particle for the time of 20 seconds be given as y. The distance travelled for the first 10 seconds is y 1 and the distance travelled for the next 10 seconds be y 2 .
Let the constant acceleration be denoted as a. Let the first 10 seconds be
denoted as t 1, the total time as t. Here,
y=y 1 +y 2

Now, using the second equation for linear motion at constant acceleration, the distance y 1 is obtained as
follows;

$${y_1} = \dfrac{1}{2}a{t_1}^2 \\\ {y_1} = \dfrac{1}{2}a{(10)^2} \\\ {y_1} = \dfrac{1}{2}a(100) \\\$$

Now, the total distance travelled by the particle in 20 seconds is given by:

$$y = \dfrac{1}{2}a{t^2} \\\ y = \dfrac{1}{2}a{(20)^2} \\\ y = \dfrac{1}{2}a(400) \\\$$

Now, the distance travelled by the particle in the seconds 10 to 20 is given as:

$${y_2} = y - {y_1} \\\ {y_2} = \dfrac{1}{2}a(400 - 100) \\\ {y_2} = \dfrac{1}{2}(300) \\\$$

Thus, the relation between y 1 and y 2 is given as follows:
y 2 =3y 1

Hence, option (B) is the correct answer.

Notes: Since, the particle is moving at a constant acceleration, the velocity is continuously increasing, which is why the distance y 2 will always be greater than y 1 . These kinematic equations of linear motion are acceptable only when there is constant acceleration and no friction. Also, change in acceleration with respect to time is called a jerk.