A body at rest is given an initial acceleration of 6.0m/s² for 20sec after which the acceleration is reduced to 4.0m/s for the next 10secs

John Perolski
August 22, 2023

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

QUESTION 64: A body at rest is given an initial acceleration of 6.0m/s² for 20sec after which the acceleration is reduced to 4.0m/s for the next 10secs.

The body maintain the speed attained for 30secs.

Draw the velocity-time graph of the motion using the information given above. 

From the graph, calculate: 

(i) Maximum speed attained during the motion; 

(ii) Total distance travelled during the first 30s; 

(iii) Average speed during the same time interval as in (ii) above

WAEC 2009e" Ans: (i) 120ms'¹ (ii) 2200m (iii) 73.33ms'¹ 

SOLUTIONS 
Initial acceleration a¹ = 6.0m/s²
Time t = 20s
Final acceleration a² = 4m/s
Time t = 10s
Constant time = 30s

(I) maximum speed attained 
v = u + at
Total time t = 20s
Maximum acceleration a = 6.0m/s²
Initial velocity u = 0
V = 0 + 6 x 20
V(1)= 120m/s

(ii) Total Distance travelled during the first 30s 

Time taken t = 20 + 10 = 30s

Total acceleration a = 6-4= 2m/s²
u = 120m/s 
t = 10s
v = 120 + 2(10)
v (2)= 120+20 = 140m/s

Total velocity, 
V = V(¹) + V(²) 
V = 120 + 140 
V = 260m/s
Total time taken t = 20 +10 = 30s

Or

Let V1 = maximum velocity after 20 sec.
V2 = maximum velocity after 30 sec.
V1 = 6 x 20 = 120m/s
V2 = (6-4) x 10 = 2 x 10 = 20m/s
VT = 120+20 = 140m/s

(ii) Total distance covered = Area of Δ + Area of trapezium after 1st 30 seconds 
Area of A = ( ½ x 20 x 120) 
= 1200m

Area of B = ½ (120 + 160) x 10 
B = 1400m

Total distance S = 1200 +1400
=2600m 

(III) Average speed = total distance/total time 
Average Speed = 2600/30
Average Speed = 86.67m/s Answer