Question

3 problems, physics

O-8.66 points SerCP10 2 P025 O Ask Your Teacher My Notes A steam catapult launches a jet aircraft from the aircraft carrier John C. Stennis, ghving it a speed of 155 mi/h in 2.40 s (a) Find the average acceleration of the plane. m/s2 (b) Assuming the acceleration is constant, find the distance the plane moves Need Help? Read it O-6.00 points SerCP10 2 P.027MLFB My Notes O Ask Your Teacher An object moving with uniform acceleration has a velocity of 13.0 om/s in the positive x-direction when its x-coordinate is 2.82 cm. If its x-coordinate 2.55 s later is-5.00 cm, what is its acceleration? cm/s Need Help? Master Read It O-.6 points SerCP10 2.P029 My Notes Ask Your Teachen A truck covers 40.0 m in 7.80s while uniformly slowing down to a final velocity of 2.20 m/s. (a) Find the truck's original speed m/s (b) Find its acceleration. m/s Need Help? Read

I only need nunber 6 and 7, sorry

Answer 1

6.

By second kinematics law,

S = u*t + 0.5*a*t^2

here,

S = displacement of object = xf - xi

S = (-5.00) - 2.82 = -7.82 cm

u = initial velocity = +13.0 cm/s

t = time interval = 2.55 sec.

a = acceleration = ??

So, by bold equation,

-7.82 = 13.0*2.55 + 0.5*a*2.55^2

a = -(7.82 + 13.0*2.55)/(0.5*2.55^2)

a = -12.6 cm/sec.

7.

By first kinematics law,

v = u + a*t

here,v = final velocity = 2.20 m/s

u = initial speed = ??

a = acceleration = ??

t = time = 7.80 sec.

So, 2.20 = u + 7.80*a eq(1)

By second kinematics law,

S = u*t + 0.5*a*t^2

here, S = distance covers = 40.0 m

So, 40.0 = u*7.80 + 0.5*a*7.80^2

40.0 = 7.80*u + 30.42*a eq(2)

By solving bold equation,

7.80*eq(1) - eq(2),

2.20*7.80 - 40.0 = (7.80*7.80 - 30.42)*a

a = (2.20*7.80 - 40.0)/30.42

a = -0.75 m/s^2

now by eq(1),

u = 2.20 - 7.80*a = 2.20 - 7.80*(-0.75)

u = 8.05 m/s

therefore answer = part-a) = u = 8.05 m/s

part-b) = a = -0.75 m/s^2

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