1. The velocity of a particle is given by v = 20t2 − 100t + 50, where v is in
meters per second and t is in seconds. Plot the velocity v and acceleration
a versus time for the first 6 seconds of motion and evaluate the velocity
when a is zero.
Solution. The acceleration is obtained by successive differentiation of v
with respect to the time. Thus,
a = 40t − 100
(1)
The velocity and acceleration are plotted against the time as shown. When
a = 0, so that
100
5
t=
=
(2)
40
2
and, the velocity becomes
5
5
v = 20( )2 − 100( ) + 50 = 125 − 250 + 50 = −75 m/s
2
2
(3)
2. The position vector of a particle moving in the x-y plane at time t = 3.60 s
is 2.76i−3.28j m. At t = 3.62 s its position vector has become 2.79i−3.33j
m. Determine the magnitude v of its average velocity during this interval
and the angle θ made by the average velocity with the x-axis.
Solution. The x- and y-components of velocity are
vx =
2.79 − 2.76
= 1.5 m/s,
3.62 − 3.60
vy =
−3.33 − (−3.28)
= −2.5 m/s
3.62 − 3.60
(4)
The total velocity becomes
v = 1.5i − 2.5j m/s
where the magnitude of v and the angle θ are
q
vy
v = vx2 + vy2 = 2.92 m/s, θ = tan−1
= −59.0◦
vx
(5)
(6)
3. A particle moves in a circular path of 0.4-m radius. Calculate the magnitude a of the acceleration of the particle (a)if its speed is constant at 0.6
m/s and (b)if its speed is 0.6 m/s but is increasing at the rate of 1.2 m/s
each second.
Solution. With the constant velocity given, we can compute the acceleration from
0.62
a=
= 0.9 m/s2
(7)
0.4
With the unit vectors en and et , the total acceleration becomes
a = 0.9en + 1.2et m/s2
where the magnitude of a is
p
a = 0.92 + 1.22 = 1.5 m/s2
1
(8)
(9)