Ask a New Question

Question

A particle travels along the x axis with a constant acceleration of +2.7 m/s2. At a time 4.2s following its start, it is at x = -108 m. At a time 5.7 s later it has a velocity of +13.0 m/s. Find its position at this later time.

Please use kinematic equations to solve.
10 years ago

Answers

since a = 2.7 m/s^2, at t=4.2, v=11.34m/s
Then its acceleration drops to (13.0-11.34)/5 = 0.332 m/s^2

So, starting at t=4.2,

x = -108 + 0.332t + 0.166 t^2
At 5.7s after the first phase, then x = -100.7
10 years ago

Related Questions

The particle travels along the path defined by the parabola y=0.5(x^2). If the component of velocity... A particle travels to the right at a constant rate of 6.5 m/s. It suddenly is given a vertical acc... 1) A particle travels between two parallel vertical walls separated by 16 m. It moves toward the opp... A particle travels between two parallel ver- tical walls separated by 39 m. It moves to- ward the... A particle travels along the x-axis so that its velocity is given by v(t)=cos3x for 0<(or equal to)t... a particle travels in a straight line ,such that for a short time 2s<t<6s its motion is described by... A particle travels along the curve c(t) = (6t^3-6, 7t^2-4). If time t is measured in seconds and if... A particle travels along the parabola y=ax^2+x+b. At what point do its abscissa and ordinate change... A particle travels between two parallel ver- tical walls separated by 12 m. It moves to- ward the op...
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use