Question

The speed of a bus increases uniformly from 15 ms per second to 60 ms per second in 20 seconds. calculate 1. the average speed 2. the acceleration 3. the distance travelled during the entire period

Answer 1

vf= 60m/s
vs = 15m/s
t=20s
a = vf-vs/t
a = 45m/s / 20s
a = 2,25 m/s²
S = vs *t + 1/2 a*t²
S = 15 m/s*20s +1/2 * 2,25m/s² * 400 s²
S = 300m +  450m
S = 750m
v average = 750m/20s
v average = 37,5 m/s

Answer 2

Answer:

Part a)

$v_{avg} = 37.5 m/s$

Part b)

$a = 2.25 m/s^2$

Part c)

$d = 750 m$

Explanation:

Part a)

Since we know that bus is moving uniformly from initial speed of 15 m/s to 60 m/s in 20 seconds

So we will have

$v_{avg} = \frac{d}{t}$

we know that

$d = \frac{v_f + v_i}{2} t$

so we have

$v_{avg} = \frac{v_f + v_i}{2}$

$v_{avg} = \frac{15 + 60}{2}$

$v_{avg} = 37.5 m/s$

Part b)

As we know that acceleration is rate of change in velocity

so it is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{60 - 15}{20}$

$a = 2.25 m/s^2$

Part c)

distance moved by the bus is given as

$d = v_{avg} \times t$

$d = 37.5 \times 20$

$d = 750 m$