71No. You cannot take a(t) as that. The motion is a linear one i.e, one dimensional along x - axis.
You can do it without graph. Some maths at hand. Basic Integration. a = dv/dt
The acceleration of a particle is given by a(t)=(3.00m/s2)-(2.00m/s3)t.
(a)Find the initial speed vosuch that the particle will have the same x-coordinate at t=5.00s as it had at t=0.
(b)What will be the velocity at t=5.00s?
(Do we have to take a as 3i-2tj??
can we solve the problem without plotting any graph ?? )
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11 Answers
71
158ok...let us assume that the particle moves only along the x-axis
then a=3-2t
v=∫a.dt
=∫(3-2t).dt
=3t-2t22
=3t-t2
Thus,at t=5s,
v=15-25=-10ms-1
how should we solve (a)??
71You carried out the integration incorrectly. Either you integrate with correct limits or use a constant of Inetgeration, that constant is something you have a great interest in.
158if we use definite integration,what are the limits?? t=0s and t=5s???
11@Anik I have just told to remember...yaad rakhne me to koi tax nehi lagata!!! :P
158@epsilon arey bhai...mai kahaa tax de raha hu...:P
@vivek can you please explain your answer?? i couldnt understand how you got 56ms-1 as initial velocity
anyways...you are correct...you got the right answer
1161Given: a(t)=3-2t
→dv/dt=3-2t
→v=3t-t^2+c1
Similarly
S=3t^2/2-t^3/3+c1t+c2
As far the condition
S(t=0) is eaual to S(t=5)
Putting the values we get C1=5/6
And initial velocity will be v(t=0)
which will be 5/6 m/s