contestada

The brakes are applied to a moving vehicle, causing it to uniformly slow down. While slowing, it moves a distance of 40.0 m in 7.45 s to a final velocity of 1.50 m/s, at which point the brakes are released.
What was its initial speed (in m/s), just before the brakes were applied?
What was its acceleration (in m/s2) while the brakes were applied? (Assume the initial direction of motion is the positive direction. Indicate the direction with the sign of your answer.)

Respuesta :

bharathparasad577

The acceleration is 1.03m/s^2 and initial velocity is 6.1735m/s

As it is given that motion is in positive direction and in one dimension, we will use the formulas of motion in one dimension

It is given to us that while slowing vehicle moves 40m

time taken to halt is 7.45s

final velocity is 1.5 m/s

We are required to find initial speed and acceleration

Using the equation v = u + at

v=1.5 , t = 7.45

1.5 = u  + a x 7.45

u + 7.45a= 1.5

u =1.5 -7.45a

Now using the equation v^2 -u^2 =2aS

1.5 x 1.5 -u^2 = 2 x a x 40

2.25 - u^2 = 80a`

2.25- 80a = u^2

Now squaring for u in first equation and equation with second ,we get:

(1.5 -7.45 a)^2 = 2.25 -80a

2.25 + 55.50a^2 -22.35a =2.25 -80a

55.5a^2+57.65 a =0

a(55.5a +57.65)=0

a= 57.65 /55.5 =1.03 m/s^2

u = 1.5 - 7.45 x 1.03

   =-6.1735 m/s

negative sign implies that initial speed is in opposite direction

#SPJ9

Otras preguntas