7 [In this question all distances are measured in metres.] A particle P is moving along the x-axis. At time t seconds, P is at the point with coordinates (x,, 0), whercx_P=8-10t+1/3t^3 Find, in terms of t,(a)the velocity of P at time t seconds,(2)(b)the acceleration...
A particle P is moving along the x-axis. At time /=0. the particle is at the origin O and is moving with speed 2 m s- in the direction Ox. At time / seconds, where / 0. the acceleration of Pis4e- m s-2 directed away from O.Find the velocity of P at time r seconds. ...
A particle P is moving along the positivex-axis in the direction of x increasing.When OP = x metres, the velocity of Pisvm s-! and the acceleration of P is(4k^2)/((x+1)^2)ms^(-2), wherek sa poiiveconstant. At x= 1, v= 0.Deduce that v cannot exceed 2k. ...
A particle P of mass m is moving along the x-axis Ox in the direction of x increasing. At timer seconds, the velocity of P is vm s-1 and the acceleration is -(a2 + 2)m s-2, where a is a constant.At time r = 0, P i at O and it speed i Um s-. At time t= T, v...
【题目】 A particle P moves along the x-axis. At time / = 0. P passes through the origin moving in the positive x direction. At time r seconds, the velocity of P is vm s-' and OP = x metres. The acceleration of P is(25 - x)Given that the maximum speed of Pis 12 m s...
The derivative of position is velocity and the derivative of velocity is acceleration. Remember to use the chain rule when taking the derivative of e^(2t).Because position is x(t)=e^(2t)-e^t, then velocity isv(t)=2e^(2t)-e^t and acceleration is a(t)=4e^(2t)-e^t....
1.The position function X(t) of a particle moving along the x-axis is x=-6t2+4,with x measured in meters and t in seconds. A)At what time B)where does the particle momentarily stop? 2.If the position of a particle is given by x=-5t3+20t,where x is in meters and t is...
A particle P is moving along a straight line. At time t=0, the particle is at a point A and is moving with velocity 8\ m s^(-1) towards a point B on the line, where AB=30\ m. At time t\ seconds (where t≥q 0), the acceleration of P is (2-2t)\ ms^(-2) in the ...
Now, the question need you to find out the position of the particle when its acceleration is zero.Henceif a=00=12-6tt=2sThen, you need to substitute t=2s into the original equation x = 6.0 t^2 − 1.0 t^3.x=6.0(2^2)-1.0(2^3) =24-8 =16mHence, when the position is 16m...
A particle is moving along the y-axis, according to the equation y(t)=ln t-2sin t for 0< t≤q (π )2. Find the position of the particle at the instant that the acceleration is zero. ( ) A. -1.661 B. 1.277 C. 8.415 D. 9.817 相关知识点: 试题来源: 解析 A 反馈 收藏 ...