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某质点做匀加速直线运动，经过时间t速度由变为，位移大小为x。则在随后的时间内，质点的位移大小为，则k的数值为（　　） A．2 B．3 C．4 D．5

Mar 19, 2024

某质点做匀加速直线运动，经过时间t速度由变为，位移大小为x。则在随后的时间内，质点的位移大小为，则k的数值为（　　） A．2 B．3 C．4 D．5

The student's question seems to be incomplete, as the velocities and time intervals are not provided in the question. However, I will demonstrate how to approach a problem like this using the kinematic equations for uniformly accelerated motion.

Solution

Assume initial velocity is

v_i

, final velocity after time

t

v_f

, and acceleration is

a

. The displacement

x

in time

t

is given by the equation:

x = v_i t + \frac{1}{2} a t^2

For the subsequent time

t

, the initial velocity for this period is

v_f

from the previous period. The displacement

x'

in the next time

t

is given by:

x' = v_f t + \frac{1}{2} a t^2

To find the ratio

k = \frac{x'}{x}

, we need to express

x'

in terms of

x

. Since

v_f = v_i + at

, we can substitute

v_f

in the equation for

x'

x' = (v_i + at)t + \frac{1}{2} a t^2

Simplify the expression for

x'

to find

k

x' = v_i t + a t^2 + \frac{1}{2} a t^2 = x + \frac{1}{2} a t^2

. Thus,

k = \frac{x + \frac{1}{2} a t^2}{x}

Without the specific values for

v_i

v_f

, and

a

, we cannot determine the exact value of

k

. However, if we had the numerical values, we could calculate

k

and select the correct multiple-choice answer

[Insert the correct letter choice here once the values are provided]

Key Concept

The key concept involved in this question is the use of kinematic equations for uniformly accelerated motion to relate displacement, velocity, and acceleration over equal time intervals.

Explanation

The explanation involves applying the kinematic equations to two consecutive time intervals to find the ratio of the displacements in those intervals. Without specific numerical values, we cannot determine the exact value of

k