Question

An Object Moving With 6M Per Second Execute An Acceleration 2M /S2 In Next3S How Much Distance It Covered (S=Ut+1/2At2

Answer 1

As initial velocity U = 6m/s

As acceleration A = 2m/s²

As time T = 3

Let's find the distance S = UT + $\huge\frac{1}{2}$ A(T)²

S = 6 × 3 + $\huge\frac{1}{2}$ × 2 × 3²

S = 18 + 9

distance S = 27m

Answer 2

Given,

The velocity with which an object is moving = 6 m/s

Acceleration of the object = 2 m/s^2

Time taken to accelerate = 3 seconds

To find,

The distance covered by the object in this 3 seconds.

Solution,

We can simply solve this numerical problem by using the following process:

As per the principles of linear motion;

If the velocity of a body is u, acceleration is a, and time taken to accelerate is t, then the total distance covered in time t is represented as:

s = ut + 1/2at^2

Now, according to the question;

The distance covered by the object in this 3 seconds

= ut + 1/2at^2

= (6 m/s)(3 s) + 1/2×(2 m/s^2)(3 s)^2

= 18 m + 9 m

= 27 meters

Hence, the distance covered by the object in the 3 seconds is equal to 27 meters.