Kinematics Problem Solving

Problem solving 1

A particle moves with equal position with respect to time:

r (t) = 3t2 – 2t + 1

with t in seconds and r in meters.

Determine:
The speed of the particle at t = 2 second

Problem solving 2

A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.

a = (v-u)/t

a = (46.1 m/s – 18.5 m/s)/(2.47 s)

a = 11.2 m/s2

d = vi*t + 0.5*a*t2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2

d = 45.7 m + 34.1 m

d = 79.8 m

Problem solving 3

With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.

First, find speed in units of m/s:

v2 = u2 + 2 a s

(0 m/s)2 = u2 + 2 (-9.8 m/s2) (91.5 m)

0 m2/s2 = u2 – 1793 m2/s2

1793 m2/s2 = u2

vi = 42.3 m/s

Now convert from m/s to mph:

vi = 42.3 m/s * (2.23 mph)/(1 m/s)

vi = 94.4 mph

Problem Solving 4

The following figure depicts the relationship between the speed and timing of the moving body straight.

Velocity after 5 seconds is …

first calculate a

a = u-v/t

= (0-6)/2

= -3 m/s2

and then calculate v

v = u + a t

= 6 + (-3) 5

= 6 – 15 = -9 m/s

Problem solving 5

A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.

v2 = u2 + 2 a s

(0 m/s)2 = vi2 + 2 (-9.8 m/s2) (2.62 m)

0 m2/s2 = u2 – 51.35 m2/s2

51.35 m2/s2 = u2

vi = 7.17 m/s