Answer to Question #110586 in Classical Mechanics for atta

Question

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

Answer 1

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

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 $t\in [1,4]$ s.
2. The total distance is: $d = d_1 + d_2 + d_3$ , where:

$d_1 = \int_0^4(5t^2 - 10t) dt\approx26.7\\$

$d_2 = \int_4^{14}40 dt = 400\\$

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

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

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