Tutorial 2
Question 1
ai) average acc is taking the velocity at t=4.7s minus the velocity at t=0 and dividing by 4.7s.
ii) The greatest instantaneous acc can be found by finding the point where the gradient is the steepest. Using a ruler and tilting it until you touch one point at the steepest gradient. Draw that line until you cover the whole graph for greater accuracy. Hence the gradient is taking the triangle where opp is 100 kmh-1 which is 23.6 ms-1 divide by the time taken the x axis ie. the line covers the whole graph.
iii) Estimate the area using trapezium formula. Area enclosed from t=2s to t=3s.
b) Construct a line from the peak to the x axis in a time of 3.7s . Since it is uniform deceleration the distance is area of a triangle. Calc the area enclosed.
Question 2
Draw the velocity-time graph of the driver. The area under the graph gives the distance the driver takes to stop hence the safe distance .Note during reaction time of 0.25s the car continue to move at constant speed.
Question 3
a) The s-t graph is a tedious process. You need to sum the area at every 2 s to obtain displacement . Key points to remember is that the area is negative below the time axis and positive above the time axis. At v = o at t=4s is the turning point of the s-t graph. The sequence of s at t =2,4,6,8,10 and 12s. The displacement points are -40, -50 , -40 ,0 , -50 and -70m. Note :1. when velocity is negative , the gradient is negative.
2. The velocity is constant at t = 6-8s, and t= 10-12s, the gradient is constant on the s-t graph.
b) The a-t graph is just the gradient of the v-t graph. Take note of positive and negative gradient which is fall above and below the x axis of the a-t graph. Graph looks like steps...
Question 4.
You need equation of motion to solve this very important question.
One involving s and time will be s=ut + 0.5at2. This equation is simplified if the motion is considered from the top where u = 0. Hence s = 0.5at2.
Consider first time t1 when the bottom edge reach the marker, s = 1.8 m .
The second time t2 is when the top edge reach the marker, s = (1.8 + L)m
Writing the two equations : 1.8 =0.5at'2 and 1.8+L =0.5at"2 but t"= t'+0.172s.
Question 5
First decide on the master arrow , down positive.
a) s = + 2.0 m , a= +9.81, and u = 0. Use v= u + at
b)repeat a) but find v using v2 = u2 + 2as.
c) The second leg involves upward motion , hence change master arrow to , up positive.
s = +1.5m, v = 0 ms-1, and a = -9.81ms-2 using v2= u2 + 2as find u
d) repeat the data in c) but use v= u + at, the u = +value
e) Drawing the v-t graph for the whole motion require one master arrow , down positive.
Velocity is down (+) at the start and up (-) after rebound. At the time of contact with the ground the direction changes from a large positive value to a negative value in a short time , hence the line is vertical. The gradient of this graph is constant as the acceleration due to gravity is g = + 9.81ms-2 throughout.
The area enclosed in the triangle is larger for the positive displacement s=2.0m and smaller for the negative displacement s = -1.5 m(up).
Question 6
Graph work required with some patience and perserverance.
a) The area under a-t graph is the v value .
Take the time markings to be 1s, 2s, 3s and 4s for easy reference.
Graph A, the area is increasing and the value v is negative. The increase in area is greater at t = 0 to 1 s and slower from t= 1 to 2s . Similarly t= 2-3 and 3-4s. Draw a curve which is steeper at both ends. If you calc the area and plot the graph will be more accurately drawn but not necessary.
Graph B requires the same method. The a = 0 at two occasions t=1 and 3s. Hence there is a turning point on the v-t graph at those two times. At t= 2 s the positive area and negative area cancels out, hence the v = 0 .The area is + and then - to cancel out and hence v= 0 .Ignore the axes of graph at the top of page, the negative sine graph is the v-t graph of graph B.
Question 7
Equation 1 for stone 1: 200 = 0.5 at2
Equation 2 for stone 2 : 200 = 20t' + 0.5at'2.
t=6.39s and t'= 4.66s.
Comparing t - 6.39s and 4.66+1s = 5.66s, the second stone will hit the ground first.
Hence the first stone is still above ground.
s= 0.5 (9.81)(5.66)2 = 157 m . Hence it's 200-157 = 43 m above the ground.
Question 8
a) Follow A , positive to the right
b) i) graph B crosses the axis , v = 0 ms-1
ii) A and B intersect, the v is the same and hence they are closest
iii) numerical value gets smaller , ignore sign . Graph B , first quarter and last quarter section .
iv) positive acc is positive gradient of graph , at 1st , 2nd and 3rd quarter section.
The graph will be used for discussion as it is quite complex .
