Question

Question 13 (2 points) A particle moving with a constant acceleration has a velocity of 20 cm/s when its position is x 10 cm. Its later is r--30 cm. What is the acceleration of the particle? position 7.0 s -7.3 cm/s -8.9 cm/s2 -11 cm/s2 -15 cm/s2 -13 cm/s2

if you could be as detailed as possible, that would be great. I'm having touble knowing which equations to use, showing the equation and why would also be appreciated.

Answer 1

Well, These type of questions can be easly solved by equations of linear motion. there is mainly three equations-

V = u+at

V^2 = u^2+2aS

S = ut+(1/2)at^2

Here u = initial velocity of particle

V = Final velocity of particle

S = displacement coverd by the particle

t = time taken

a = acceleration of the particle.

Now you have to check question carefully that which datas are given and what we have to find..choose equations accorngly.

Question no 13:

Here given Initial velocity u =20 cm/s = 0.20 m/s

Displacement S = X2-X1 = -30-10 = -40cm = -0.40m

time t =7 s

And acceleration a = we have to find.

so now see here u, S and t is given and we have to find 'a' so clearly seen that equation no 03 have all these terms so use third equation.

S = ut+0.5at^2

-0.40= 0.20*7 + 0.5*a*7^2. = -0.073 m/s^2 = -7.3 cm/s^2

option A is correct.

Question no 14.

u = 30 m/s

V = 50 m/s

S = 180 m

t = ?

There is no direct equation for implimentation of these three terms so first of all we have to calculate acceleration using 2nd equation.

V^2 = u^2 +2aS

a = (V^2 -u^2)/2S

= (50^2-30^2)/(2*180) = 4.44 m/s

now from equation first

V = u+at

t = (V-u)/a

= (50-30)/4.44 = 4.50 seconds

2nd options is correct.

Question no 15:

S = 80m

t = 5 s

u = 20 m/s

a = ?

From third equation

S = ut + (1/2)at^2

80 = 20*5 + 0.5*a*5^2

a = -1.6 m/s^2 .....option first is correct.

Hope this helps you. please thumbs up?