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 I m/s.
After 10 seconds from start of observations the body was 4.5I 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 1.5I m 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 = (4t2 –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 5I 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 = I t + 3t2 –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 = t3 + 3t2 + 4t + 1.2
I .
Find :
(i) Velocity at start and after 4 seconds
(ii) Acceleration at start and after 4 seconds
1.
A.
Acceleration at start (t=0): a = 2 – 3(0) = 2 m/s^2
Velocity at start (t=0): v = 0 + 2(0) = 0 m/s
Distance from the origin at start (t=0): 0 m
B.
To determine when velocity becomes zero, set v = 0:
0 = 2 – 3t
t = 2/3 seconds
Distance from the origin when velocity is zero:
s = s(0) + v(0)*t + 1/2 a t^2
s = 0 + 0 + 1/2 (2)(2/3)^2
s = 4/9 m
2.
A.
Acceleration is the derivative of velocity:
a = d(v)/dt = 6t^2 - 2t - 2
B.
Displacement after t seconds:
s = ∫(2t^3 - t^2 - 2t + 4) dt = 1/2 t^4 - 1/3 t^3 - t^2 + 4t + C
For t = I seconds:
s = 1/2 I^4 - 1/3 I^3 - I^2 + 4I + C
Given that at t = 2 s, s = 1.5I:
1.5I = 1/2(2^4) - 1/3(2^3) - (2^2) + 4(2) + C
1.5I = 8 - 8/3 - 4 + 8 + C
C = 3.5I
3.
Position of the particle at t = 4 seconds:
s = ∫(4t^2 - 2) dt = 4/3 t^3 - 2t + C
Given s(0) = 2 and s(2) = 5I:
2 = 4/3(0)^3 - 0 + C
C = 2
5I = 4/3(2)^3 - 2(2) + 2
5I = 32/3 - 4 + 2
I = 4/3
Position at t = 4 seconds:
s = 4/3(4)^3 - 2(4) + 2
s = 85 1/3 m
4.
A.
Velocity at start (t = 0): v = d(s)/dt = 1 + 0 - 0 = 1 m/s
Acceleration at start (t = 0): a = d(v)/dt = 0 + 6 - 6(0) = 6 m/s^2
B.
To find maximum velocity, set acceleration to 0:
0 = 6 - 6t
t = 1 second
C.
Max velocity is at t = 1 second:
v = 1 + 3(1) - 2(1)^2 = 2 m/s
5.
(i)
Velocity at start (t = 0): v = d(s)/dt = 0 + 3(0)^2 + 4 = 4 m/s
Velocity after 4 seconds (t = 4): v = 3(4)^2 + 2(4) + 4 = 60 m/s
(ii)
Acceleration at start (t = 0): a = d(v)/dt = 0 + 6(0) + 2 = 2 m/s^2
Acceleration after 4 seconds (t = 4): a = 6(4) + 2 = 26 m/s^2