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. To find acceleration at the start, we set t = 0 in the equation a = 2 - 3t:
a = 2 - 3(0) = 2 m/s^2
To find the velocity at the start, we integrate the acceleration function to get the velocity function:
v = ∫(2 - 3t) dt = 2t - (3/2)t^2 + C
Given the initial velocity after 5 seconds is I m/s:
I = 2(5) - (3/2)(5)^2 + C
C = I + 25/2
So, the velocity at the start (t = 0) is:
v = 0 + 0 + I + 25/2 = I + 25/2 m/s
The initial distance from the origin at the start can be calculated by integrating the velocity function:
s = ∫(I + 25/2) dt = It + (25/2)t + D
Given the distance after 10 seconds is 4.5I m from the origin:
4.5I = I(10) + (25/2)(10) + D
D = -5I
Therefore, the distance from the origin at the start (t = 0) is:
s = 0 + 0 - 5I = -5I units
B. To find the time after start of observation in which the velocity becomes zero, set v = 0 in the velocity function:
0 = 2t - (3/2)t^2 + I + 25/2
3t^2 - 2t - 25 - 2I = 0
Solving this quadratic equation will give you the time at which velocity becomes zero. Once you have the time, substitute it into the distance equation s = It + (25/2)t - 5I to find the distance from the origin.
2.
A. To find the acceleration, differentiate the velocity function with respect to time:
a = dv/dt = 6t^2 - 2t - 2 m/s^2
B. To find the displacement after t seconds, integrate the velocity function:
s = ∫(2t^3 - t^2 - 2t + 4) dt = (1/2)t^4 - (1/3)t^3 - t^2 + 4t + D
Given the distance is 1.5I m from station A at t = 2 seconds, you can solve for the constant D and then substitute t = I to find the displacement after I seconds.
3.
Given the acceleration function, you can integrate it to get the velocity function and then integrate again to get the position function s(t). Once you have the position function, you can then find the position of the particle when t = 4 seconds.
4.
A. To find the velocity at the start, differentiate the equation of motion with respect to time:
v = ds/dt = d/dt (I t + 3t^2 - 2t^3) = I + 6t - 6t^2
To find the acceleration at the start, differentiate the velocity function:
a = dv/dt = da/dt = 6 - 12t
B. The time when the particle reaches its maximum velocity can be found by setting the acceleration equal to zero and solving for t. To find the maximum velocity, substitute the value of t into the velocity equation.
5.
(i) To find the velocity at the start, differentiate the displacement function with respect to time and evaluate at t = 0. Repeat the process to find the velocity after 4 seconds.
(ii) To find the acceleration at the start, differentiate the velocity function with respect to time and evaluate at t = 0. Repeat the process to find the acceleration after 4 seconds.