1. A body moves along a straight line and its acceleration a .which varies with t is given by
a = 2 – 3t. After 5 seconds from start of observations its velocity is observed to be 041*m/s.
After 10 seconds from start of observations the body was 4.5*041 metres from the origin.
A. Determine its acceleration, velocity and distance from the origin at the start of
observations.
B. Determine the time after start of observation in which the velocity becomes zero and its
distance from the origin.
2. A particle moves along a straight line with a velocity given by the equation
v = 2t3 – t2 – 2t + 4
where v is the velocity in m/s and t is time is seconds. When t = 2 seconds, the particle is found
to be at a distance of 1.5*041m from a station A. Determine:
A. the acceleration and
B. displacement of the particle after I seconds
3. A particle moves along a straight line with an acceleration described by the equation
a = (4t^2 –2), where a is in m/s2 and t is in seconds. When t = 0, the particle is located 2 m
to the left of origin and when t = 2 seconds, the particle is 5*041 m to left of origin.
A. Determine the position of the particle, when t = 4 seconds.
4. The equation of motion of a particle moving in a straight line is given by
s = 041*t + 3t^2 –2t^3
where s is the total distance covered from the starting point is metres at the end of t seconds.
Find :
A. the velocity and acceleration at start,
B. the time, when the particle reaches its maximum velocity and
C. the maximum velocity of the particle.
5. A particle moves along a straight line so that its displacement in metre from a fixed point is
given by, s = t^3 + 3t^2 + 4t + 1.2*041
.
Find :
(i) Velocity at start and after 4 seconds
(ii) Acceleration at start and after 4 seconds.
1.
A.
Given a = 2 - 3t, we can integrate with respect to t to find the velocity v:
v = ∫(2 - 3t) dt = 2t - 3(t^2)/2 + C
At t = 5 seconds, v = 41 m/s:
41 = 2(5) - 3(5^2)/2 + C
C = 41 + 15/2 = 46.5
So, v = 2t - 3(t^2)/2 + 46.5
Now, at t = 10 seconds, the body is 4.5*41 m from the origin:
Distance = ∫v dt = ∫(2t - 3(t^2)/2 + 46.5) dt = t^2 - (3t^3)/6 + 46.5t + D
At t = 0, the distance from the origin is 0, so D = 0.
Therefore, at t = 10 seconds:
4.5*41 = 10^2 - (3(10^3)/6) + 46.5(10)
Solving for t = 0, we get t = 0.5 seconds
Therefore, at the start of observations:
Acceleration = 2 m/s^2
Velocity = 46.5 m/s
Distance from the origin = 0 m
B.
To find the time when the velocity becomes zero, set v = 0:
0 = 2t - 3(t^2)/2 + 46.5
Solving for t, we get t = 11 seconds
The distance from the origin at t = 11 seconds is:
Distance = 11^2 - (3(11^3)/6) + 46.5(11) = 41.5 m
2.
A.
Given v = 2t^3 - t^2 - 2t + 4, we can find the acceleration by differentiating v with respect to t:
a = dv/dt = 6t^2 - 2t - 2
At t = 2 seconds:
a = 6(2^2) - 2(2) - 2 = 20 m/s^2
B.
To find the displacement of the particle after t seconds, integrate v with respect to t:
Displacement = ∫v dt = ∫(2t^3 - t^2 - 2t + 4) dt = (t^4)/2 - (t^3)/3 - t^2 + 4t + C
At t = 0, displacement from station A is 1.5*41 m, so C = 1.5*41
Therefore, the displacement after t seconds is:
Displacement = (t^4)/2 - (t^3)/3 - t^2 + 4t + 1.5*41
3.
Given a = 4t^2 - 2, we can find the velocity by integrating a w.r.t. t:
v = ∫(4t^2 - 2) dt = (4t^3)/3 - 2t + C
At t = 0, the particle is located 2 m to the left of origin, so C = -2
Therefore, the velocity of the particle is:
v = (4t^3)/3 - 2t - 2
At t = 2 seconds, the particle is 5*41 m to the left of origin:
Position = ∫v dt = ∫((4t^3)/3 - 2t - 2) dt = t^4 - t^2 - 2t + D
When t = 2, position = 5*41:
5*41 = 2^4 - 2^2 - 2(2) + D
D = 5*41 - 20
D = 1
Therefore, the position of the particle at t = 4 seconds is:
Position = 4^4 - 4^2 - 2(4) + 1 = 283 m
4.
A.
Given s = 41t + 3t^2 - 2t^3, we can find the velocity by differentiating s with respect to t:
v = ds/dt = 41 + 6t - 6t^2
At the start (t = 0):
Velocity = 41 m/s
Acceleration is found by differentiating v with respect to t:
a = dv/dt = 6 - 12t
At t = 0:
Acceleration = 6 m/s^2
B.
To find the time when the particle reaches its maximum velocity, set acceleration a = 0:
0 = 6 - 12t
t = 0.5 seconds
C.
To find the maximum velocity, substitute t = 0.5 into v:
v = 41 + 6(0.5) - 6(0.5)^2
v = 46 m/s
5.
(i)
Given s = t^3 + 3t^2 + 4t + 1.2*41, differentiate s with respect to t to find the velocity:
v = ds/dt = 3t^2 + 6t + 4
At t = 0:
Velocity at start = 4 m/s
At t = 4 seconds:
Velocity = 3(4)^2 + 6(4) + 4 = 52 m/s
(ii)
Differentiate velocity v to find acceleration a:
a = dv/dt = 6t + 6
At t = 0:
Acceleration at start = 6 m/s^2
At t = 4 seconds:
Acceleration = 6(4) + 6 = 30 m/s^2