Movement of a car A car. Of mass m-1500Kg, moves on a straight and horizontal ro

Question

Movement of a car A car. Of mass m-1500Kg, moves on a straight and horizontal road Ox. Knowing that at the date to-0, the car is in O. A. Car movement study: The variation of the speed of the car, as a function of time, is represented in the graph below: (m/s) 20 15 10 t (s) 10 20 0 40 60 1) The car goes through two phases. Specify these phases. (nature and time intervals) 2) a) Calculate the acceleration of the car during each phase. b) Write the equations giving the speed V of the car as a function of time. 3) Calculate the distance traveled by the car during each phase and deduce the total distance traveled.

Explanation / Answer

1] First phase is with uniform velocity, from t=0 to t = 40s

Second phase is with uniform retardation (negative acceleration) from t = 40s to t=60s

2] a) a = 0 m/s^2   for t < 40s

a = dv/dt = (0-20)/(60-40) = - 1 m/s^2......for 60s > t > 40s

Note: there is error in graph, on time scale, instead of 60s, 50s should be mentioned. Here I have taken 60s, but the answer may differ(-2 m/s^2).

b) v = 20 m/s for t < 40s

v = 20 - 1*(t-40) = 60 - t for 60s > t > 40s

3] distance travelled in first phase = 20*40 = 800 m

distance travelled in second phase = 20*20*0.5 = 200 m

B)1)a) F will act forward, f act backward.

b) in first phase, F-f = ma = 0

F = f = 2500N answer

2]a) f and f' act backward.

b) -f - f' = ma

-2500 - f = 1500*a , here a is the value found in second phase, having value -1 or -2 depending on the graph's time scale 60s or 50s respectively.

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