This question is “making the rounds” in Singapore schools in Feb 2012. Its a tough question because the thinking behind it is a little different than standard Kinematics problems.
Question: Tom uses a stopwatch to measure the time between 2 lamp-posts on a car journey. As the car speeds up, 2 consecutive times are 1.2s and 1.0s. Later, he finds out that the lamp-posts are 30m apart.
(a) Calculate the acceleration of the car
(b) calculate the speed at the first lamp-post.
A) His speed in the middle of the first time interval is 30m/1.2s = 25 m/s.
His speed in the middle of the second time interval is 30m/1s = 30 m/s
To find the acceleration, take change in speed over change in time. The change in speed is 5m/s. The change in time is 0.5s+0.6s = 1.1s. (the amount of time between the middle of the two time intervals). The acceleration is then 5/1.1 = 4.5 m/s^2.
B) To get the velocity at the moment he passes the first lamp post, take 25m/s to be a final velocity 0.6s later. Use: v= u+at. Solve for u, the initial velocity at the time he passes the first lamp post. 25m/s = u + 4.5(0.6s). u = 22.3 m/s
This problem is unusual for secondary level in that it involves realizing that the speed you calculate is in the middle of a time interval. That’s the key !