the acceleration of a particle is defined by the relation a=-8m/s^2. Knowing tha

Question

the acceleration of a particle is defined by the relation a=-8m/s^2. Knowing that x=20m when t=4s and that x=4m when v=16m/s, determine (a) the time when the velocity is zero, (b) the velocity and the total distance traveled when t=11s

Explanation / Answer

a=-8; x1=4; v1=16; x2=20; t2=4; You can tell that the points are given out of chronological order by noticing that if x1 were 20 and x2 were 4, the velocity at x=4 would have been negative. using the position formula (same as integrating): x2 = x1 + v1*(t2-t1) + (a/2)*(t2-t1)^2 solve for the only unknown t1 20 = 4 + 16(4-t1) - (8/2)*(4-t1)^2 => t1 = 2 V(t) = V1 + a(t-t1) 0 = 16 -8(t-2) => V=0 @ t=4 V(t) = V1 + a(t-t1) V(11) = 16 -8(11-2) => V=-56 @ t=11 Total distance is rather vague, id get clarification. Assuming it means total distance traveled between t=0 and t=11: V(t) = V1 + a(t-t1) => V(0) = 16-8(0-2)= V(0)=32 x0 + v0*t1 + (a/2)*t1^2 = x1 x0 + 32*2 - (8/2)*2^2 = 4 => x0 = -44 (x @ t=0) at t=11, x= 4+16(11-2) - (8/2)*(11-2)^2 = -176 so at t=0, x=-44. at t=4, x=20 and the particle has stopped. at t=11, x=-176 so assuming the t=0 to t=11 assumption about the question is right, distance traveled is: (44+20)+(20+176)=260 or if you took it to mean the change in position between t=0 and 11, 176-44=132

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