1) A car starts from rest and moves along the X axis with constant acceleration `5 m/s^2` for `8` seconds. If it then continues will constant velocity, what distance will the car cover in `12`...
1) A car starts from rest and moves along the X axis with constant acceleration `5 m/s^2` for `8` seconds. If it then continues will constant velocity, what distance will the car cover in `12` seconds since it started from the rest?
2) A motorcyclist drives from A to B with a uniform speed of 30 km per hour and returns back with a speed of 20 km per hour. Find its average speed.
3) The velocity-time graph given below shows the motion of a cyclist. Find: 1) its acceleration 2) its velocity 3) the distance covered by the cyclist in a second.
4) An object starting from rest travels 20 meter in first 2 seconds and 160 meters in next 4 seconds, what will be the velocity after 7 seconds from the start?
When the acceleration `a` is constant, the speed `V` is equal to `V(t)=V_0+at,` where `V_0` is the initial speed. The distance traveled is `D(t)=V_0 t+(a t^2)/2.`
1) The final speed is `V_1=V_0t+at=0+5*8=40(m/s).` The distance of accelerated motion is `0+(5*8^2)/2=160(m),` the distance of constant speed motion is `V_1*(12-8)=40*4=160(m).` So the total distance is 320m.
4) (The same formulas): probably at both periods the acceleration is constant and is the same.
For the first period `D_1=20=0+(a*2^2)/2,` so `a=10(m/s^2),` and `V_1=0+a*2=20(m/s).`
The data for the second period give us nothing new, although there is no contradiction: `D_2=160=V_1*4+(10*4^2)/2=80+80` is true.
So after `7` seconds the speed will be `a*7=70(m/s).`
2) This is a different but simple story. The average speed is by definition (the total distance traveled) / (total time spent). In our case it is
`(2 AB)/((AB)/30 + (AB)/20) = 2/(1/30+1/20)=(2*20*30)/(20+30)=24(m/s).`
Here AB denotes the distance between A and B.