PRACTICAL PHYSICS

I have tried my hands on this and have still not be able to come to a perfect conclusion, Now i want all those nairaland first class to help me out on this. Here is the question. all I need is basic and concise explanation, give me the values and say whether to convert km/h to secs(m/sec) stating the hows and whys.Would the graph be curved, how do i find the slop and intercept?Also tell me how tofind the speed and acceleration from the graph.

The speedometer of a motor-car showed the following speeds in km/h at the end of successive intervals of 3 secs:
38.9, 52.8, 64.4,73.6, 80.9, 86.1, 90.0, 92.5. Plot the velocity-time graph in convenient units and find from the graph the disatance in km covered while the speed increased from 80 to 90km/h. and the accelerations when the speeds are 50,70 and 90km/h

Have you been able to plot the velocity-time graph?

You don't have to plot a graph to know that the acceleration of an object at a given speed is zero,a=dv/dt.It'll be easier to use m/s instead of km/s considering a small time interval of 3s.Using a scale of 4cm rep 5m/s on the y-axis and 2cm rep 3s on the x-axis,the distance covered from 80-90km/h is approx 0.013km.the graph is an irregular one,more like a series of straight lines,

Thanks for the response, actually this is what i did, but still,


Speed(Km/h) 38.9 52.8 64.4 73.6 80.9 86.1 90.0 92.5

Since the time is at successive intervals of 3secs, i assume the rest would be 3s, 6s, 9s, 12s, 15s, 18s,,21s,and 24secs. therefore using speed as the y-axis and Time as the x-axis, i should have something like this:

Speed(Km/h) 38.9 52.8 64.4 73.6 80.9 86.1 90.0 92.5

Time(secs) 3 6 9 12 15 18 21 24.

The problem now is: can i work with these values to plot the graph(plotting the graph is compulsory) or do i need to change km/h to m/sec? if i work with these values i would only get a curved velocity-time graph with only three best fit point(only three values are straight-line) how do i then determine the slope to calculate the acceleration from the graph? Although i agree the graph is irregular, but someone please help me out on this, i appreciate any response.

jarsofclay:

Thanks for the response, actually this is what i did, but still,


Speed(Km/h) 38.9 52.8 64.4 73.6 80.9 86.1 90.0 92.5

Since the time is at successive intervals of 3secs, i assume the rest would be 3s, 6s, 9s, 12s, 15s, 18s,,21s,and 24secs. therefore using speed as the y-axis and Time as the x-axis, i should have something like this:

Speed(Km/h) 38.9 52.8 64.4 73.6 80.9 86.1 90.0 92.5

Time(secs) 3 6 9 12 15 18 21 24.

The problem now is: can i work with these values to plot the graph(plotting the graph is compulsory) or do i need to change km/h to m/sec? if i work with these values i would only get a curved velocity-time graph with only three best fit point(only three values are straight-line) how do i then determine the slope to calculate the acceleration from the graph? Although i agree the graph is irregular, but someone please help me out on this, i appreciate any response.

First thing first!

Put all those variables in the same unit. I'll advise you convert km/hr to m/s.

Its important you always work with the same units in any question involving calculations,as i posted earlier its easier to use m/s rather than km/h except you're comfortable working in micro-hours,if you plot the graph as a series of lines-as i described earlier,converting the velocity to m/s the velocites will fall into different lines,you can calculate the acceleration by finding the slope of the line,at 50km/h,the acceleration from my graph is 1.3m/s^2,at 70km/h,the acceleration is 0.83m/s^2 etc.