Question

A passenger bus is travelling 28.0 m/s to the right when the driver applies the brakes. The bus stops in 5.00 s. What is the acceleration of the bus as it comes to a stop?

Answer 1

change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right

Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s

Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right

Answer 2

Answer:  - 5.60 m/s^2

Explanation:

Speed of passenger bus = 28.0 m/s to the right  

Time took for the bus stops = 5.00 s.

What is the acceleration of the bus as it comes to a stop = x

Step 1:

Calculate the change in velocity  

d = v2 (Final velocity)  - v1 (Initial Velocity)

d= 0 – 28

d= -28m/s  

Acceleration=  Change in velocity / time taken

Acceleration = -28 m/s / 5

Acceleration = - 5.60 m/s^2 to the right

Note: Do not forget the units, otherwise they make cost you marks!