Question 156: Motion in 1D

Question

A particle moving along x-axis has acceleration f, at time t, given by f = f
0
$$\left( {1 - {t \over T}} \right),$$ where f
0
and T are constants.The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle's velocity (v
x
) is

Accepted Answer

f = f
0
$$\left( {1 - {t \over T}} ight)$$
When f = 0 then,
0 = f
0
$$\left( {1 - {t \over T}} ight)$$
$$ \Rightarrow $$ $$\left( {1 - {t \over T}} ight)$$ = 0
$$ \Rightarrow $$ t = T
Also f = $${{dv} \over {dt}}$$
$$ 	herefore $$ $$\int\limits_0^{{v_x}} {dv = \int\limits_{t = 0}^{t = T} {fdt} } $$
v
x
= $${\int\limits_{t = 0}^{t = T} {{f_0}\left( {1 - {t \over T}} ight)dt} }$$
= $${{f_0}\left[ {\left( {t - {{{t^2}} \over {2T}}} ight)} ight]_0^T}$$
= $${{f_0}\left( {T - {{{T^2}} \over {2T}}} ight)}$$
= $${{f_0}\left( {T - {T \over 2}} ight)}$$
= $${1 \over 2}$$ f
0
T

Answer

$${1 \over 2}$$ f
0
T
2

Answer

f
0
T
2

Answer

$${1 \over 2}$$ f
0
T

Answer

f
0
T

Motion in 1D

A particle moving along x-axis has acceleration f, at time t, given by f = f 0 $\left( {1 - {t \over T}} \right),$ where f 0 and T are constants.The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle's velocity (v x ) is

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