Question

<p>The position function x(t) of a particle moving along an x axis is x = 6.00 - 7.00t2, with x in meters and t in seconds. (a) At what time and (b) where doesthe particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin? Graph x versus t for the range -5 s to +5s. (e) To shift the curve rightward on the graph, should we include the term +20t (denote 1) or the term -20t (denote 0) in x(t)? (f) Does that inclusion increase(denote 1) or decrease (denote 0) the value of x at which the particle momentarily stops?<br /> <br /> <br /> <br /></p>

Answer 1

x = 6.00 - 7.00t²

diferntial will give velocity.

dx/dt = 0 - 7 * 2 t

when velocity is zero particle stops.

∴dx/dt = 0 = - 14t

a) at time t = 0 the particle momentarily stops.

At t=0

x =6.00 - 7.00 (0²) = 6.00 meter

b) at position x = 6.00 meter the particle has stopped.

c) origin is given as x =0.

∴ x = 6.00 - 7.00t²

0 = 6 - 7t²

7t² = 6 ------> t² = 6/7 or t = ±0.93 secnds

so at t = + 0.93s and t = -0.93s