4. A train that is travelling at 30 m s–1 slows at a uniform rate to rest in 40 seconds. Calculate the acceleration and stopping distance.
Solution u = 30 m s–1
v = 0 m s–1 t = 40 s
s = ut + 1
_ 2 at2
s = (30)(40) + 1
s = 1200 – 600 s = 600 m
5. A motorbike is accelerating uniformly as it passes two checkpoints that are 45 m apart. The time taken to pass between the checkpoints is 5 seconds and the velocity of the bike as it passes the first checkpoint is 4 m s–1. Calculate the acceleration of the bike and its velocity as it passes the second checkpoint.
Solution
Consider the 1st checkpoint as being s = 0. u = 4 m s–1 s = 45 m
t = 5 s a = ?
s = ut + 1
_ 2 at2
45 = (4)(5) + 1
_ 2 (a)(5)2
45 = 20 + 12.5a 25 = 12.5a 2 m s–2 = a u = 4 m s–1 a = 2 m s–2 t = 5 s v = ?
You may now complete Exercise 2B (page 20). Graphs
We can illustrate velocity and acceleration with respect to time in two different types of graph: distance–time graphs and speed–time graphs.
LEAVING CERTIFICATE PHYSICS 15
_ 2 (–0.75)(40)2
a = v – u
______ t
a = 0 – 30
________ 40
a = −0.75 m s–2
v = u + at v = (4) + (2)(5) v = 14 m s–1
6. A plane moves from rest and accelerates uniformly prior to take-off. It covers a displacement of 0.5 km in 10 seconds. Calculate the (i) acceleration, (ii) velocity at take-off and (iii) distance covered during the 10th second.
(i) u = 0 m s–1 s = 500 m
s = ut + 1
_ 2 at2
500 = (0)(10) + 1
500 = 0 +50a 10 m s−2 = a
(ii) v = u + at v = (0) + (10)(10) v = 100 m s–1
(iii) By calculating the distance covered in 10 seconds and subtracting the distance covered in 9 seconds, you can determine the difference in distances and find the distance travelled in the 10th second.
Distance travelled in 9 seconds: s = ut + 1
_ 2 at2
s = (0)(9) + 1 s = 405 m
_ 2 (10)(9)2
Distance travelled in 10 seconds: 500 m (from question)
Therefore distance travelled in 10th second: 500 m – 405 m = 95 m
