Variable acceleration (1D)

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ap-calculus-bc 2012 Q89 Find displacement/position by integrating velocity
A particle moves along a line so that its acceleration for $t \geq 0$ is given by $a ( t ) = \frac { t + 3 } { \sqrt { t ^ { 3 } + 1 } }$. If the particle's velocity at $t = 0$ is 5, what is the velocity of the particle at $t = 3$ ?
(A) 0.713
(B) 1.134
(C) 6.134
(D) 6.710
(E) 11.710
csat-suneung 2011 Q17 4 marks True/false or multiple-statement verification
Point P starts from the origin and moves on a number line. The velocity $v ( t )$ of point P at time $t ( 0 \leqq t \leqq 5 )$ is as follows. $$v ( t ) = \begin{cases} 4 t & ( 0 \leqq t < 1 ) \\ - 2 t + 6 & ( 1 \leqq t < 3 ) \\ t - 3 & ( 3 \leqq t \leqq 5 ) \end{cases}$$ For a real number $x$ with $0 < x < 3$, let $f ( x )$ be the minimum among:
  • the distance traveled from time $t = 0$ to $t = x$,
  • the distance traveled from time $t = x$ to $t = x + 2$,
  • the distance traveled from time $t = x + 2$ to $t = 5$.
Which of the following are correct? Choose all that apply from $\langle$Remarks$\rangle$. [4 points]
$\langle$Remarks$\rangle$ ㄱ. $f ( 1 ) = 2$ ㄴ. $f ( 2 ) - f ( 1 ) = \int _ { 1 } ^ { 2 } v ( t ) d t$ ㄷ. The function $f ( x )$ is differentiable at $x = 1$.
(1) ㄱ
(2) ㄴ
(3) ㄱ, ㄴ
(4) ㄱ, ㄷ
(5) ㄴ, ㄷ
csat-suneung 2017 Q10 3 marks Find velocity/speed by differentiating position
A point P moving on the coordinate plane has position $( x , y )$ at time $t$ $(t > 0)$ given by $$x = t - \frac { 2 } { t } , \quad y = 2 t + \frac { 1 } { t }$$ What is the speed of point P at time $t = 1$? [3 points]
(1) $2 \sqrt { 2 }$
(2) 3
(3) $\sqrt { 10 }$
(4) $\sqrt { 11 }$
(5) $2 \sqrt { 3 }$
csat-suneung 2017 Q12 3 marks Compute total distance traveled over an interval
The velocity $v ( t )$ at time $t ( t \geq 0 )$ of a point P moving on a number line is $$v ( t ) = - 2 t + 4$$ What is the distance traveled by point P from $t = 0$ to $t = 4$? [3 points]
(1) 8
(2) 9
(3) 10
(4) 11
(5) 12
csat-suneung 2019 Q27 4 marks Determine velocity at zero acceleration
The position $x$ at time $t$ ($t \geq 0$) of a point P moving on a number line is $$x = - \frac { 1 } { 3 } t ^ { 3 } + 3 t ^ { 2 } + k \quad ( k \text{ is a constant} )$$ When the acceleration of point P is 0, the position of point P is 40. Find the value of $k$. [4 points]
csat-suneung 2020 Q27 4 marks Two-particle comparison problem
Two points P and Q move on a number line. Their positions $x _ { 1 } , x _ { 2 }$ at time $t ( t \geq 0 )$ are $$x _ { 1 } = t ^ { 3 } - 2 t ^ { 2 } + 3 t , \quad x _ { 2 } = t ^ { 2 } + 12 t$$ Find the distance between points P and Q at the moment when their velocities are equal. [4 points]
csat-suneung 2021 Q14 4 marks Compute total distance traveled over an interval
The velocity $v ( t )$ of a point P moving on a number line at time $t ( t \geq 0 )$ is $$v ( t ) = 2 t - 6$$ If the distance traveled by point P from time $t = 3$ to time $t = k$ ($k > 3$) is 25, what is the value of the constant $k$? [4 points]
(1) 6
(2) 7
(3) 8
(4) 9
(5) 10
csat-suneung 2023 Q20 4 marks Compute total distance traveled over an interval
The velocity $v ( t )$ and acceleration $a ( t )$ of a point P moving on a number line at time $t$ ($t \geq 0$) satisfy the following conditions. (가) When $0 \leq t \leq 2$, $v ( t ) = 2 t ^ { 3 } - 8 t$. (나) When $t \geq 2$, $a ( t ) = 6 t + 4$.
Find the distance traveled by point P from time $t = 0$ to $t = 3$. [4 points]
csat-suneung 2024 Q10 4 marks Two-particle comparison problem
At time $t = 0$, two points P and Q simultaneously depart from the origin and move on a number line. Their velocities at time $t$ ($t \geq 0$) are respectively $$v_1(t) = t^2 - 6t + 5, \quad v_2(t) = 2t - 7$$ Let $f(t)$ denote the distance between points P and Q at time $t$. The function $f(t)$ increases on the interval $[0, a]$, decreases on the interval $[a, b]$, and increases on the interval $[b, \infty)$. Find the distance traveled by point Q from time $t = a$ to time $t = b$. (Here, $0 < a < b$) [4 points]
(1) $\frac{15}{2}$
(2) $\frac{17}{2}$
(3) $\frac{19}{2}$
(4) $\frac{21}{2}$
(5) $\frac{23}{2}$
csat-suneung 2026 Q11 4 marks True/false or multiple-statement verification
There is a point P that starts from the origin at time $t = 0$ and moves on a number line. For a real number $k$, the velocity $v ( t )$ of point P at time $t$ ($t \geq 0$) is $$v ( t ) = t ^ { 2 } - k t + 4$$ Which of the following in are correct? [4 points]
ᄀ. If $k = 0$, then the position of point P at time $t = 1$ is $\frac { 13 } { 3 }$. ㄴ. If $k = 3$, then the direction of motion of point P changes once after departure. ㄷ. If $k = 5$, then the distance traveled by point P from time $t = 0$ to $t = 2$ is 3.
(1) ᄀ
(2) ᄀ, ᄂ
(3) ᄀ, ᄃ
(4) ㄴ, ㄱ
(5) ᄀ, ᄂ, ᄃ
jee-main 2005 Q3 Inverse/implicit relationship between position and time
The relation between time $t$ and distance x is $\mathrm{t} = a\mathrm{x}^2 + \mathrm{bx}$ where a and b are constants. The acceleration is
(1) $-2abv^2$
(2) $2bv^3$
(3) $-2av^3$
(4) $2av^2$
jee-main 2012 Q2 Find acceleration from position or velocity
The distance travelled by a body moving along a line in time $t$ is proportional to $t^{3}$. The acceleration-time $(a, t)$ graph for the motion of the body will be
(1) [graph 1] (2) [graph 2] (3) [graph 3] (4) [graph 4]
jee-main 2019 Q1 Velocity as a function of position
A particle is moving with speed $v = b \sqrt { x }$ along positive $x$ - axis. Calculate the speed of the particle at time $t = \tau$ (assume that the particle is at origin at $t = 0$ )
(1) $b ^ { 2 } \tau$
(2) $\frac { b ^ { 2 } \tau } { \sqrt { 2 } }$
(3) $\frac { b ^ { 2 } \tau } { 2 }$
(4) $\frac { b ^ { 2 } \tau } { 4 }$
jee-main 2020 Q21 Inverse/implicit relationship between position and time
The distance $x$ covered by a particle in one dimensional motion varies with time $t$ as $x ^ { 2 } = a t ^ { 2 } + 2 b t + c$. If the acceleration of the particle depends on $x$ as $x ^ { - n }$, where $n$ is an integer, the value of $n$ is $\_\_\_\_$
jee-main 2020 Q21 Find velocity/speed by differentiating position
A particle is moving along the $x$-axis with its coordinate with time $t$ given by $x ( t ) = 10 + 8 t - 3 t ^ { 2 }$. Another particle is moving along the $y$-axis with its coordinate as a function of time given by $y ( t ) = 5 - 8 t ^ { 3 }$. At $t = 1 \mathrm {~s}$, the speed of the second particle as measured in the frame of the first particle is given as $\sqrt { v }$. Then $v$ (in $\mathrm { m s } ^ { - 1 }$) is $\_\_\_\_$.