Derive third equation of motion v² - u² = 2as by graphical method. b) A 1000 Kg vehicle moving with a speed of 20 m/s is brought to rest in a distance of 50 m: 1) Find the acceleration. 2) Calculate the unbalanced force acting on the vehicle.
Answers
Explanation:
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Answer:
a) From the graph, we can say that
The total distance travelled, s is given by the Area of trapezium OABC.
Hence,
S = ½ (Sum of Parallel Sides) × Height
S=(OA+CB)×OC
Since, OA = u, CB = v, and OC = t
The above equation becomes
S= ½ (u+v) × t
Now, since t = (v – u)/ a
The above equation can be written as:
S= ½ ((u+v) × (v-u))/a
Rearranging the equation, we get
S= ½ (v+u) × (v-u)/a
S = (v2-u2)/2a
Third equation of motion is obtained by solving the above equation:
v2 = u2+2aS
b) refer to the notes
c) Since we that acceleration is -4 m/s²
Now,
F=ma
F=1000×−4
=−4000N
