Question

a car moving with a velocity of 54km/h accelerate uniformly at the rate of 2m/s².Calculate the distance travelled from place where acceleration bagans to where the velocity reaches 72km/h and the time taken to cover the distance​

Answer 1

ANSWER: 234.4 m and 2.5 seconds

Step-by-step explanation:

1. List what we have.

`v_f = 72 km/h=20 m/s`

`v_i = 54 km/h = 15 m/s`

`a = 2 m/s^2`

2. Find time first.

`t = (v_f - v_i)/(a)`

`t = (20 - 15)/(2)`

`t= 2.5` seconds

2. Then find the distance.

`s = v_i t (1)/(2) a t^2`

`s = (15)(2.5)(1)/(2) (2)(2.5)^2`

`s = 234.4` m

btw, I'm not good at physics AT ALL so this is just me guessing and trying to help

Answer 2

let's first convert km/hr into m/s
54 km/hr : 15m/s
72 km/hr : 20m/s
Applying :
v
2
−
u
2
=
2
a
S
v
2
−u
2
=2aS
20
×
20
−
15
×
15
=
2
×
2
×
S
S
=
175
/
4
=
43.75
m
20×20−15×15=2×2×S
S=175/4=43.75m