1) A car moves at 10 m/s for a period of 5 seconds then slows to 5 m/s for the next 10 seconds. Draw a graph showing the complete motion. What is the speed of the car after 2.5 seconds? What is its speed after 8 seconds? What is the total distance covered over 15 seconds? Over the first 5 seconds?

Solution:

Graph: On a velocity –time scale will show the time axis (horizontal) labeled from 0 to about 15 s. The vertical scale will show velocity from 0 to 15 m/s at intervals graduated say in single m/s. First part of graph is a

*horizontal line*from edge of vertical scale at 10 m/s to a time of 5 s on horizontal scale. The graph then slopes from the coordinate (5s, 10 m/s) down to (15s, 5 m/s) in a straight line inclination.
From this graph, the speed (velocity) after 2.5s is 10 m/s – which is constant over the

*first 5 seconds*.
Beyond 5 s the car is decelerating at the rate of: a = -(10 m/s – 5 m/s)/ 10 sec = 5 m/s/ 10s = 0.5 ms

^{-2}. So that the acceleration reduces the car’s speed for 3 s hence: v = u + at = (10 m/s) + (-0.5 ms^{-2}) (3s) = 10 m/s – 1.5 m/s = 8.5 m/s.*area under graph*:

Constant velocity portion = L x W = (10m/s) (5s) = 50m (Distance covered over first 5 s)

^{-2}) (10s) = 25 m so total = 75m

We may alternately use the kinematic equation: v

^{2}= u^{2}+ 2as
Or: v

^{2}- u^{2}= 2as
Where v = 5 m/s and u = 10 m/s

So: s = (v

^{2}- u^{2})/ 2a = [(5 m/s)^{2}- (10 m/s)^{2 }] / 2 (-0.5 ms^{-2}) = (-75 m^{2}/s)^{2 }/ 2a
Or s = (-75 m

^{2}/s)^{2 }/ 2( -0.5 ms^{-2}) = 75m2) A go Kart starts off with an initial uniform velocity of 2 m/s for 5 s. It then accelerates to 3 m/s in 1 second and sustains that for 3 seconds. What is the distance it covers?

Solution:

The distance over the

*uniform velocity interval*is: d1 = (2 m/s) (5 s) = 10 m
The distance covered over the

*acceleration interval*is: d2 = ½ at^{2}
Where a = v/ t = ( 3 m/s)/ 1 s = 3 ms

^{-2}
So: d2 = ½ (3 ms

^{-2})( 3s ) 2 = 27/ 2 m = 13.5 m
So the total distance = d1 + d2 = 10m + 13.5 m = 23.5 m

This can also be done using the kinematic equation: s = ut + ½ at

^{2}
In the form: s = ut1 + ½ at2

^{2}^{}^{}
Where u = 2 m/s and t1 = 5 s, while a = 3 ms

^{-2}and t2 = 3 s
So: s = (2 m/s) (5s) + ½ (3 ms

^{-2})( 3s )^{2 }= 23.5 m
3) A car is traveling along a level road at 50 km/h, then it speeds up to 80 km/h over a period of 5 minutes. Find the acceleration in m s

^{-2}and the distance covered in the time, in m if it maintains the final velocity for 1 minute. (Take 1000 m = 1 km, 1 h = 3600 s)*Solution*:

Acceleration = (v – u)/ t = (80 km/h - 50 km/h)/ 5 min.

But 5 mins. = 5/ 60 = 1/12 h

So, acceleration = (80 km/h - 50 km/h)/ (1/12 h) = 30 km/h/ 1/12 h = 360 km/ h

^{2}
Find the acceleration in m s

^{-2}:
a = 360 km/ h

^{2}= 360, 000 m/ (3600s)^{2}= 0.027 m s^{-2}
Distance s covered in the time in meters is s = ½ at

^{2}(Note time t = 5 min = 300 s)
s = ½ (0.027 m s

^{-2}) (300s)^{2}= 1 215 m^{}Add the distance resulting from the final velocity (80 km/h) sustained for one minute (60s)

Then, since 80 km/h = 80 000m/ 3600 s = 22.2 m/s

Then over 60s: d = 22.2 m/s (60s) = 1332 m

Then total distance = s + d = 1215 m + 1332 m = 2547 m

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