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Xelga [282]
3 years ago
7

A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult

. Assuming the acceleration is constant, how far did it travel during those 2.0 s?
Physics
1 answer:
Dvinal [7]3 years ago
5 0

Answer:

Explanation:

Given

time taken t=2\ s

Speed acquired in 2 sec v=42\ m/s

Here initial velocity is zero u=0

acceleration is the rate of change of velocity in a given time

a=\frac{v-u}{t}

a=\frac{42-0}{2}=21\ m/s^2

Distance travel in this time

s=ut+0.5at^2

where  

s=displacement

u=initial velocity

a=acceleration

t=time

s=0+\0.5\times 21\times (2)^2

s=42\ m

so Jet Plane travels a distance of 42 m in 2 s                                

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The student soon loses his balance and falls backwards off the board at a velocity of 1.0 m/s. Assuming momentum is conserved in
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Answer:

v2 = 27.3m/s

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Assuming forward as positive.

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v2 = ( (64 + 5.94)×1.4 + 64×1.0)/5.94

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