The diagram above (Fig. 1.17) shows a velocity- time graph representing the motion of a car. Find the total distance covered during the acceleration and retardation periods of the motion

Physics Jamb, Waec, Neco Past Questions & Answers on Speed, Velocity and Acceleration

QUESTION 35: The diagram above (Fig. 1.17) shows a velocity- time graph representing the motion of a car. Find the total distance covered during the acceleration and retardation periods of the motion. A. 75m B. 150m C. 300m D. 375m 

JAMB 1993 Ans: 75m

SOLUTIONS

Total distance s ,= Area of shape

Area = [A + B + C]

A = ½(b+h)

B = b x h

C = ½(b x h)

Solving for A; Total distance during Acceleration 

b = 10-0 = 10

h = 10-0 = 10

Therefore 

A = ½ (10 x 10)

A = 100/2 

A = 50m

Solving for B (Not necessary) total distance during Constant velocity 

b = 40-10 = 30

h = 10-0 =10

Therefore 

B = b X h

B = 30 x 10 

B = 300m

Solving for C ; Total distance during deceleration 

C = ½(b xh)

b = 45-40 = 5

h = 10-0 =10

Therefore 

C = ½ ( 5 x 10)

C = ½ x 50

C = 50/2

C = 25m

Hence total distance s = distance during acceleration A + distance during deceleration C

S= A + C

S =50m + 25m

S = 75m ANSWER