Question #110586

a particle P moves in a straight line . the velocity vms-1 at time t s is given by

v=5t(t-2) for 0<=t<=4

v=k for 4<=t<=14

v=68-2t for 14<=t<=20

where k is constant

1) find k.

2)sketch the velocity time graph for 0<=t<=20

3)find the set of values of t for which the acceleration of P is positive

4)find the total distance travelled by P in the interval 0<=t<=20

Expert's answer

  1. In order to get continuous velocity function it is required 5t(t2)=k5t(t-2) = k at t=4t = 4. Thus k=40.k = 40.

The velocity time graph for 0<=t<=20:

  1. The acceleration of P is positive when velocity increases. According to the graph, it is true for t[1,4]t\in [1,4] s.
  2. The total distance is: d=d1+d2+d3d = d_1 + d_2 + d_3 , where:

d1=04(5t210t)dt26.7d_1 = \int_0^4(5t^2 - 10t) dt\approx26.7\\

d2=41440dt=400d_2 = \int_4^{14}40 dt = 400\\

d3=1420(682t)dt=204d_3 = \int_{14}^{20}(68-2t) dt = 204\\

Thus, d=26.7+400+204=630.7d = 26.7+400+204=630.7 m.


Answer. 1) k = 40; 2) see graph; 3) t in [1,4]s; 4) d = 630.7m.


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