Question

particle moves along. straight line with speed v=t-2ms find out distance traveled in first 4 seconds​

Answer 1

Answer:

initial velocity , u = 0m

Final velocity , v = 2m

Time taken , t = 2 second

Displacement , d=?

Now acceleration = V-U / T

= 2-0 / 2

=1m/s

Now according to equation of motion ,

V^2 - U^2 = 2as

2^2 - 0^2 = 2*1*s

4-0 = 2s

4=2s

4/2= s

2m= S

Therefore displacement in first four seconds = 2m

Answer 2

GiveN :

• v = (t - 2) m/s

To FinD :

• Distance travelled in 4s

SolutioN :

⇒v = dx/dt

⇒t -2 = dx/dt

⇒dx = (t - 2).dt

Integrate both sides

⇒∫dx = ∫(t - 2).dt

⇒x = ∫t.dt - ∫2.dt

⇒x = t²/2 - 2t + C

Put t = 4

⇒x = 4²/2 - 2(4) + C

⇒x = 16/2 - 8 + C

⇒x = 8 - 8 + C

⇒x = (0 + C)m.

• Where C is constant of integration