There is a car traveling at 30 ms-2 and it passes a police car.
2 seconds after the car passes the police car the police car starts accelerating at 4ms-2 until he overtakes the car.
however 8 seconds after passing the police car, the car accelerates at 2ms-2.
how far will the police car travel until it overtakes the car.
2 seconds after the car passes the police car the police car starts accelerating at 4ms-2 until he overtakes the car.
however 8 seconds after passing the police car, the car accelerates at 2ms-2.
how far will the police car travel until it overtakes the car.
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I think you have a typo. 30 ms-2 be 30 ms. I have done the rest of the problem assuming this.
Drawing a velocity-time graph might help you visualize what's going on. The graph should start at the point the car passes the police car (t=0 at that point). Let's observe both vehicles separately.
Car: It is traveling at 30 m/s for 8 sec and then accelerates at 2 ms-2
So, on the V-T graph, there should be a horizontal line at v =30 for 8 seconds and then a line from T = 8 onwards with a slop of 2.
Police: For the first 2 seconds, the velocity is zero. Then the police accelerates at 4 ms-2 for the rest of the time.
Graphically, there is a horizontal line on the x-axis for two seconds and then a line from T = 2 onwards with a slope of 4.
Now, we will find equations for both of the vehicles velocities. If we keep the same starting point, we will have to use piece-wise functions so to keep things simple, we will the graph from the time the car starts to accelerate (T=8) because then we are only dealing with sloped lines. N.B: This point does not become T = 0, it is still T = 8. When we determine the slope-intercept equations for the velocities of both vehicles, the intercepts (B) will be the velocities of the respective cars at T=8.
Car: V = 2T + 30
Polive: V = 4T +24
However, we need the distance the police car travels so we need to convert these velocity equations into displacement ones. So, integrate the velocity equation and figure out how far both cars had traveled at T=8. In case you don't know Calculus, integrating the velocity equations above will get you the following equation: df = (1/2)aT^2 + v0T + d0
(the f means final and the 0 means initial).
Finding the distance traveled up to T=8.
Car: (30 m/s)(8 s) = 240 m
Police: (1/2)(24 m/s)(8 s) = 72 m
Car: df = T^2 +30T + 240
Police: df = 2T^2 +24T+ 72
Now set the distances equal to each other and solve for T. T = 16.3 s.
Use that in either equation (although the police's equation is more logical since you want the distance the police traveled) to solve for df.
Df = 995 m.
So the police car traveled 0.995 km before overtaking the car.
Drawing a velocity-time graph might help you visualize what's going on. The graph should start at the point the car passes the police car (t=0 at that point). Let's observe both vehicles separately.
Car: It is traveling at 30 m/s for 8 sec and then accelerates at 2 ms-2
So, on the V-T graph, there should be a horizontal line at v =30 for 8 seconds and then a line from T = 8 onwards with a slop of 2.
Police: For the first 2 seconds, the velocity is zero. Then the police accelerates at 4 ms-2 for the rest of the time.
Graphically, there is a horizontal line on the x-axis for two seconds and then a line from T = 2 onwards with a slope of 4.
Now, we will find equations for both of the vehicles velocities. If we keep the same starting point, we will have to use piece-wise functions so to keep things simple, we will the graph from the time the car starts to accelerate (T=8) because then we are only dealing with sloped lines. N.B: This point does not become T = 0, it is still T = 8. When we determine the slope-intercept equations for the velocities of both vehicles, the intercepts (B) will be the velocities of the respective cars at T=8.
Car: V = 2T + 30
Polive: V = 4T +24
However, we need the distance the police car travels so we need to convert these velocity equations into displacement ones. So, integrate the velocity equation and figure out how far both cars had traveled at T=8. In case you don't know Calculus, integrating the velocity equations above will get you the following equation: df = (1/2)aT^2 + v0T + d0
(the f means final and the 0 means initial).
Finding the distance traveled up to T=8.
Car: (30 m/s)(8 s) = 240 m
Police: (1/2)(24 m/s)(8 s) = 72 m
Car: df = T^2 +30T + 240
Police: df = 2T^2 +24T+ 72
Now set the distances equal to each other and solve for T. T = 16.3 s.
Use that in either equation (although the police's equation is more logical since you want the distance the police traveled) to solve for df.
Df = 995 m.
So the police car traveled 0.995 km before overtaking the car.