Question

40. Displacement of a truck moving along x-axis is
given by x = at + Bt2 - yt3, where t is time and a,
B and y are positive constants. Velocity of truck
when its acceleration is zero will be

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Answer 1

The velocity of truck is,  $v=a+\frac{B}{(3y-B)} -3y\frac{1}{4(9y^{2}+B^{2} -6yB) }$ when its acceleration is zero.

Given,

$x=at+Bt^{2} -yt^{3}$

a, B and y-constants.

To find,

velocity of truck when its acceleration is zero.

Solution:

Mathematical tool called differentiation will be used here.

Velocity is equal to the differentiation of the position of an object with respect to time.

$v=\frac{dx}{dt}\\v=\frac{d(at+Bt^{2}-yt^{3}) }{dt} \\v=a+2Bt-3yt^{2} .$

The equation of velocity came as:

$v=a+2Bt-3yt^{2}$.

Acceleration is equal to the rate of change of velocity.

It is equal to the differentiation of the velocity of an object with respect to time.

$acceleration=\frac{dv}{dt} \\acceleration=\frac{d(a+2Bt-3yt^{2} )}{dt} \\acceleration=1+2B-6yt.$

The equation of acceleration came as:

$acceleration=1+2B-6yt.$

According to the question,

$acceleration=0\\1+2Bt-6yt=0\\1+2Bt=6yt\\1=6yt-2Bt\\1=2t(3y-B)\\t=\frac{1}{2(3y-B)} .$

Acceleration will be 0 at time,  $t=\frac{1}{2(3y-B)}$ .

Velocity at this time will be:

$v=a+2Bt-3yt^{2} \\\v=a+2B{\frac{1}{2(3y-B)} }-3y\frac{1}{2(3y-B)} ^{2} \\v=a+\frac{B}{(3y-B)} -3y\frac{1}{4(9y^{2}+B^{2} -6yB) }$.

Velocity of truck, $v=a+\frac{B}{(3y-B)} -3y\frac{1}{4(9y^{2}+B^{2} -6yB) }$.

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Answer 2

Answer:

When a truck's acceleration is zero, it is moving at its velocity v = a + B/(3y - B) - 3y 1/4(y^2 + B^2 - 6yB).

Explanation:

Given,

x + at + Bt^2 - yt^2

the variables a, B, and y.

the place,

truck's speed when its acceleration is zero.

Solution:

Here, a mathematical technique called differentiation will be applied.

The difference in an object's position with regard to time is what is referred to as its velocity.

v = dx/dt

v = d(at + Bt^2 - yt^3)/dt

v = a + 2Bt - 3yt^2

The velocity equation is expressed as:

v = a + 2Bt - 3yt^2

The rate of change in velocity is the same as acceleration.

It is equivalent to differentiating the speed of an item relative to time.

acceleration = dv/dt

acceleration = d(a + 2Bt - 3yt^2)

acceleration = 1 + 2Bt - 6yt

The equation of acceleration came as:

acceleration = 1 + 2B - 6yt

The acceleration equation was:

acceleration = 0.

1 + 2Bt - 6yt = 0

1 + 2B = 6yt

1 = 6yt - 2B

1 = 2t(3y - B)

t = 1/2(3y - B)

In response to the query,

At time, acceleration will be 0.

t = 1/2(3y - B)

The velocity at this moment will be:

v = a + 2Bt - 3yt^2

= a + 2B 1/2(3y - B) - 3y 1/2(3y-B)^2

v = a + B/(3 - yB) - 3y 1/4(y^2 + B^2 - 6yB)

truck's rate of velocity.

v = a + B/(3y - B) - 3y 1/4(y^2 + B^2 - 6yB)

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