Question

A body traveling at a uniform retradation of 25ms-1 has its velocity reduced to half the initial value in traveling a distance of 540m calculate the initial velocity and the time interval to travel the distance 540m​

Answer 1

Given :

Retardation of body = 25m/s²

Distance covered by car = 540m

velocity of body reduced to half the initial value.

To Find :

Time of journey and initial velocity of body.

Solution :

■ Here body is moving with constant retardation it means velocity of body is decreasing with time. Therefore acceleration will be negative.

Let initial velocity of body be u. Therefore final velocity will be u/$.$

Applying third equation of kinematics;

• v² - u² = 2as

v denotes final velocity

u denotes initial velocity

s denotes distance

a denotes acceleration

By substituting the given values;

➠ v² - u² = 2as

➠ (u/4)² - u² = 2(-25)(540)

➠ -15u²/16 = -27000

➠ u² = 28800

➠ u = 169.7 m/s

★ Time of journey :

Applying first equation of motion;

⭆ v = u + at

⭆ 169.7/4 = 169.7 + (-25)t

⭆-25t = -127.27

⭆ t = 5.09s

Answer 2

Answer:

Given

☄retardation= 25 m/sec²

☄distance covered= 540 m

☄velocity reduced to half the initial velocity.

To Find

inital Velocity

time interval to travel the distance 540 m

Solution

★Since thier is a uniform retardation so the velocity is decreasing and hence the acceleration will be negative.

According to the third equation of kinematics-

$\bold{ {v}^{2} = {u}^{2} + 2as}$

where,

v= final Velocity

u= inital Velocity

a= Acceleration

s= distance travelled

let us assume inital Velocity as 'u' and final velocity as 'u/2 '

put the given value in the above equation of kinematics-

(u/2)² = u² +2 x (-25) x 540

u²/4 = u² - 27,000

u²-4u²/4= -27000

-3u²/4= -27000

u²= 27000 x 4/3

u² = 36000

u = √ 36000

u = 189.7 m/sec

According to 1st equation of kinematics-

$v= u + at$

189.7/ 2 = 189.7 + (-25) x t

94.85 = 189.7 -25 t

94.85= 25 t

t= 3.794 seconds