A car traveling at constant velocity of 20m/s passes a stationary police car and 15 seconds later the police car accelerates uniformly at 4m/s to reach a speed of 30m/s with which it continues. a) draw a velocity time graph. b) find where the second car overtakes the first car. c) find the distance up to which the second car overtakes the first car
Solution.
"v_c=20m\/s;"
"t_c=15s;"
"v_{p0}=0m\/s;"
"v_p=30m\/s;"
"a_p=4m\/s;"
"a)"
"b) S_c=v_c\\sdot t_c=20\\sdot15=300m;"
"a_p=\\dfrac{v_p-v_{p0}}{t_p}\\implies t_p=\\dfrac{v_p-v_{p0}}{a_p};"
"t_p=\\dfrac{30-0}{4}=7.5s;"
"x_{c0}=S_c+v_ct_p=300+20\\sdot7.5=450m;"
"x_c=450+20t;"
"x_p=30t;"
"x_c=x_p;"
"450+20t=30t;"
"10t=450;"
"t=45s;"
"c)x_c=450+20t=450+20\\sdot45=1350m;"
Answer: "b)t=45s;"
"c) x_c=1350m."
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