## Tuesday, September 10, 2013

### Applied Algebra Problem Solutions (1)

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.

Total distance covered over 15 s is area under graph:

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

The area of the decelerating portion: d = ½  bh =   ½ (0.5  ms-2) (10s) = 25 m so total = 75m

We may  alternately use the kinematic equation: v2  =  u2   + 2as

Or:  v2  -   u2   =   2as

Where v = 5 m/s and u = 10 m/s

So:  s = (v2  -   u2  )/ 2a =  [(5 m/s)2  - (10 m/s)2 ] / 2 (-0.5   ms-2) =  (-75 m2/s)2  / 2a

Or s =  (-75 m2/s)2  / 2( -0.5   ms-2)  = 75m

2) 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 at2

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 at2

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/ h2

Find the acceleration in  m s-2  :

a = 360 km/ h2   =   360, 000 m/  (3600s) 2 =  0.027 m s-2

Distance s covered in the time in meters is s  =  ½ at2   (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