A bus moves from rest with a uniform acceleration of 2m/s(square) for the first 10s. It then accelerate at a uniform rate of 1m/s(square) for another 15s, it continues at constant speed for 70s and finally comes to rest in 20s by uniform deceleration.Draw the velocity time graph using the information above.

Question

Please I need an instant answer.

Answer 1

v(0)=0
v(10)=1/2 a t^2=1/2 *2*100=100m/s
so draw a straight line from 0,0 to 10,100
v(15)=100+1/2 a (t-10)^2 =100+ 1/2 (1)(25=125m/s
so plot the point (15,125)
then keep speed a constant horizontal line until time=70+15
plot the point (105, 125)
finally velocity at time125, v=0. Plot that point.

Answer 2

answer

Answer 3

V=u+at
=0+2×10
=20ms-¹
V1= 20+1×15=35ms-¹
So the total distance traveled
Area of OAG =½×10×20=100m
Area of trapezium GABF
=½(20+30)×15=412.5m
Area of FBCE =35×70=2450m
Area of triangle ECD
=½×20×35=350m

Total distance = 100+412.5+2450+350=3312.5m

Answer 4

3312.5

Answer 5

Ifedinna