Question

A body moves with an initial velocity of 40m/s and it accelerates uniformly until it attains a velocity of 80m/s. It then continues at that velocity for some time and then decelerates uniformly to rest. The total time taken for the journey is 60s and the total distance travelled is 3200m. If the time spent accelerating is half of travelling at constant velocity, calculate the acceleration

Answer 1

Answer:

a=0.44ms^-2

Step-by-step explanation:

s=ut+1/2at^2

3200=40ms^-1 ×60+1/2×a×60×60

3200-2400=1/2×a×60×60

800×2/60×60=a

a=4/9

a=0.44ms^-2

correct me if I'm wrong

Answer 2

Answer:

5m/$s^{2}$

Step-by-step explanation:

the initial velocity, v1= 40m/s

the velocity accelerated to, v2= 80m/s

total time= 60s

total distance= 3200m

time taken for acceleration ta= 1/2 time taken for constant velocity tc

ta=$\frac{1}{2}$tc ,tc=2ta

draw the velocity-time graph, as shown below

total distance= area of velocity-time graph

3200=(1/2)(40+80)ta + (2ta*80) + 1/2(80)(60-3ta)

3200=60ta + 160ta + 2400 - 120ta

800 = 100ta

ta = 8s

the acceleration is the slope between time t=0 and t=ta=8s

acceleration,a=$\frac{80-40}{8-0}$=5m/$s^{2}$