Question

5.
A train is moving on a straight track with a speed of 70.0km/h [E]. It slows down at a rate
of 2.00m/s2 [W] in a distance of 35.0m. Determine
a.The final speed of the train. b.How much time the train took to slow down to this speed.

Answer 1

Answer:

t = 1.659s

Explanation:

We can use the kinematics equations to solve this questions:

v = u + at

$v^{2} = u^{2} +2as$

where v = Final Velocity, u = initial velocity, a = acceleration, t = time, s = displacement

a) Given information from the question,

u = $\frac{70km}{h} =\frac{(70*1000)m}{(1*3600)s} = 19.444m/s$ (Convert km/h to m/s first)

a = $2m/s^{2}$

s = 35m

Now we can substitute these values into the 2nd kinematics equation to find v, final velocity.

$v^{2} =(19.444)^{2} +2(2)(35)\\v=\sqrt{(19.444)^{2} +2(2)(35)} \\v= 22.761m/s (5.sf)$

b) Now we have the final velocity, we can substitute the values into the first kinematics equation to find t , the time taken.

v = u + at

22.761 = 19.444 + 2t

2t = 22.761 - 19.444

t =$\frac{22.761-19.444}{2}$

t = 1.659s