A race car moves such that its position fits the reiationship x = (6.0 m/s)t + (

Question

A race car moves such that its position fits the reiationship x = (6.0 m/s)t + (0.60 m/s^3)t^3 where x is measured in meters and t in seconds. A plot of the car's position versus time is which of the following? In 1865, Jules Verne proposed sending men to the Moon by firing a space capsule from a 220-m-long cannon with final speed of 10.97 km/s. What would have been the unrealistically large acceleration experienced by the space travelers during their launch? (A human can stand an acceleration of 15g for a short time.) Compare your answer with the free-fall acceleration, 9.80 m/s^2. An object moves with constant acceleration 3.60 m/s^2 and over a time interval reaches a final velocity of 13.6 m/s.

Explanation / Answer

1. (a) second option i.e. (b) in which the x limits goes to 180 m/s

as x(m) = 165.6 m at 6.0 sec

(b) velocity = dx/dt = 6 + 1.80t2

at 0.40 sec instantaneous velocity = 6.288 m/s

at 0.20 sec instantaneous velocity = 6.072 m/s

at 0.10 sec instantaneous velocity = 6.018 m/s

2. we know that

vf^2 = v0^2 + 2 a d

vf=final vel = 10,970m/s
v0=initial velocity = 0
a = accel
d=dist = 220m

(10970m/s)^2 = 0 + 2 x a x 220m

a = 273502 m/s2

this is 273502 m/s2 / 9.81m/s2g = 27880 g

3.

a) Vf^2 = Vi^2 + 2ad

=> 184.96 = 46.24+ 7.2 d

=> d = 19.267 m .............Ans.

b) here the distance = displacement = 19.267 m .............Ans.

c) Vf^2 = Vi^2 + 2ad

here the displacement is same = 19.267 m .............Ans.

d) the total distance here will be different , split the questio into 2 parts :

1) when it starts at - 6.8 m/s then stops to change direction

here 0 = 6.8^2 + 2ad

=> d = - 6.8^2 / -2(3.60) = 6.423 m

and the distance to return to its starting point = 6.423 m

2) the distance it passes from starting pt till its velocity becomes 13.6 m/s = 19.267 m ( just as (a) )

so the total distance = 2x6.423 + 19.267 = 32.111 m........Ans.

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