(a) Calculate the displacement of the ball when it reaches B from A. (2)
(b) Calculate the rate of change of velocity in this time ( 0-10 sec)? (2)
(c) What is the velocity of the ball in the 4th second of its motion?
Answers
Answer:
Uniform Acceleration
The acceleration of a body is said to be uniform if its velocity changes by equal amounts in equal intervals.
Non-Uniform Acceleration
The acceleration of a body is said to be non-uniform if its velocity changes by unequal amounts in equal intervals of time.
Average velocity<
< v_{avg}> = \frac{\int_{0}^{t}vdt}{\int_{0}^{t}dt}
Average acceleration
< a_{avg}> = \frac{\int_{0}^{t}adt}{\int_{0}^{t}dt}
Illustration:
A particle moves with a velocity v(t) = (1/2)kt2 along a straight line. Find the average speed of the particle in time T.
Solution:
< v_{avg}> = \frac{1}{T}\int_{0}^{T}v(t)dt = \frac{1}{2T}\int_{0}^{T}kt^{2} dt = \frac{1}{6}kT^{2}
Illustration:
A particle having initial velocity is moving with a constant acceleration 'a' for a time t.
(a)Find the displacement of the particle in the last 1 second.
(b)Evaluate it for u = 2 m/s, a = 1 m/s2 and t = 5 sec.
Solution:
(a) The displacement of a particle at time t is given s = ut + 1/2at2
At time (t - 1), the displacement of a particle is given by
S' = u (t-1) + 1/2a(t-1)2
So, Displacement in the last 1 second is,
St = S - S'
= ut + 1/2 at2 – [u(t-1)+1/2 a(t-1)2 ]
= ut + 1/2at2 - ut + u - 1/2a(t - 1)2
= 1/2at2 + u - 1/2 a (t+1-2t) = 1/2at2 + u - 1/2at2 - a/2 + at
S = u + a/2(2t - 1)
(b) Putting the values of u = 2 m/s, a = 1 m/s2 and t = 5 sec, we get
S = 2 + 1/2(2 x 5 - 1) = 2 + 1/2 x 9
= 2 + 4.5 = 6.5 m
Illustration:
Position of a particle moving along x-axis is given by x = 3t - 4t2 + t3, where x is in meters and t in seconds.
(a)Find the position of the particle at t = 2 s.
(b)Find the displacement of the particle in the time interval from t = 0 to t = 4 s.
(c)Find the average velocity of the particle in the time interval from t = 2s to t=4s.
(d)Find the velocity of the particle at t = 2 s.
Solution:
(a) x(t) = 3t - 4t2 + t3
=> x(2) = 3 x 2 - 4 x (2)2 + (2)3 = 6 - 4 x 4 + 8 = -2m.
(b) x(o) = 0
X(4) = 3 x 4 - 4 x (4)2 + (4)3 = 12 m.
Displacement = x(4) - x(0) = 12 m.
(c) < v > = X(4)X(2)/(4-2) = (12-(-2))/2 m/s = 7 m/s
(d) dx/dt = 3 - 8t + 3t2
v(2) (dx/dt)2 = 3 - 8 x 2 + 3 x (2)2 = -1m/s
Illustration:
Two trains take 3 sec to pass one another when going in the opposite direction but only 2.5 sec if the speed of the one is increased by 50%. The time one would take to pass the other when going in the same direction at their original speed is
(a) 10 sec (b) 12 sec
(c) 15 sec (d) 18 sec
Solution:
Using the equation,
t = d/vr
We have,
3 = d/v1+v2
2.5 = d/1.5v1+v2
Solving we get,
v1 = 2d/15 and v2 = d/5
When they are going in same direction,
vr = v2 – v1 = d/15
Thus, t = d/vr = d/(d/15) = 15 s
From the above observation we conclude that, option (c) is correct.
Explanation:
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