Let after a time t1 sec M/C attains a velocity of 20 m/s
For M/C :-
By v = u+at
=>20 = 0 + 2.5 x t1
=>t1 = 20/2.5 = 8 sec
Let the M/C travel s1 distance in 8 sec
=> By s = ut + 1/2 at^2
=>s1 = 0+ 1/2 x 2.5 x 8 x 8 = 80m
Similarly for car:-
=> 30 = 0 + 1.5 x t2
=>t2 = 30/1.5 = 10 sec
&s2 = 1/2 x 1.5 x 10 x 10 = 75m
(a) Let after t sec they both are side by side
=>80 + 20t = 75 + 30t
=>5 = 10t
=>t = 0.5 sec
Answer #2:
Bike
Vf = a*t
20 = 2.5 * t
t = 8 s
Car
Vf = a*t
30 = 1.5 * t
t = 20 s
Do not do any more motion equations
Just draw the graph and watch.
The area under a v vs t graph = distance
Graph bike starts at the origin. At 8 seconds, v = 20. Draw a line from (0,0) to (8,20). Then draw a line from (8,20) to the right ( parallel to x-axis)
You drew a triangle.
Area = ½ base * height
Area under line from (0,0) to (8,20) = 80 m
Area from there on= 20*t
d bike = 80 + 20t
Graph car starts at the origin. At 20 seconds, v = 30. Draw a line from (0,0) to (20,30). Then draw a line from (20,30) to the right ( parallel to x-axis)
Area = ½ base * height
Area under line from (0,0) to (20,30) = 300 m
Area from there on= 30*t
d car = 300 + 30t
a) After what time will the motorbike and the car again be side by side?
When they have traveled the same distance, d bike = d car.
80 + 20t = 300 + 30t
10 t = 220
t = 22 seconds
(b) What is the greatest distance that the motorbike is in front of the car?
Since the bike is accelerating faster than the car for the first 8 seconds, I bet the bike is ahead of the car the greatest distance at t = 8 seconds.
Before 8 seconds, the car is moving 15m each second, and the bike is moving 10m each second. The car is moving away from the bike. After 8 seconds, the bike is moving 20m each second and the car is still moving 15m each second. ( remember, d = ½ v *t, since we are finding the area of a triangle. ½ v = average velocity)
time…….bike area……car area…….distance apart
7……… ½ *7*20……… ½ *7*30………..35
8………½ *8 *20……….½ *8 * 30………..40
9………½ *8 * 20 + 20…½ *9 * 30………..35