Question

a body possesses an initial velocity of 5 m/s and moves with a uniform acceleration of 2 ms-2. calculate the distance moved by the body in 4s.​

Answer 1

▶ Solution :

Given :

▪ Initial velocity of body = 5mps

▪ Acceleration of body = 2m/s^2

▪ Time interval = 4s

To Find :

▪ Distance covered by body in the given interval.

Concept :

✒ Here acceleration is constant throughout the whole journey so we can apply equation of kinematics directly.

✒ Third equation of kinematics is given by

☞ S = ut + (1/2)at^2

• S denotes distance
• u denotes initial velocity
• a denotes acceleration
• t denotes time interval

Calculation :

→ S = (5×4) + (1/2)×2×(4)^2

→ S = 20 + 16

→ S = 36m

Answer 2

$\underline{\underline{\tt{\purple{Given\:that-}}}}$

• A body possesses an initial velocity of 5 m/s.
• And moves with a uniform acceleration of 2 m/s².
• Also given that time is 4 sec.

$\underline{\underline{\tt{\purple{To\:Find-}}}}$

Distance moved by the body in 4 sec.

$\underline{\underline{\tt{\purple{Solution-}}}}$

The body possesses an initial velocity of 5 m/s. Means, the initial velocity or the starting velocity i.e. u of the body is 5 m/s.

And accelerates with a speed of 2 m/s².

(Acceleration i.e. a of the body is 2 m/s².)

Time i.e. t is 4 sec (given).

Using the SECOND EQUATION OF MOTION

$\huge{\boxed{\sf{s=ut+\frac{1}{2}at^2}}}$

We have u = 5 m/s, t = 4 sec, a = 2 m/s².

Substitute these values in boxed formula,

$\rightarrow \rightarrow\:\sf{s=5(4)+\frac{1}{2}(2)(4)^2}$

$\rightarrow \rightarrow\:\sf{s=20+(1)(16)}$

$\rightarrow \rightarrow\:\sf{s=20+16}$

$\rightarrow \rightarrow\:\sf{\red{s=36}}$

Therefore,

The distance travelled by the body is 36 m.