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

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

Answer Link

angeraha angeraha · Answer 2 · 2021-09-19T19:47:27+02:00

ANSWER: 234.4 m and 2.5 seconds

EXPLANATION:

1. List what we have.

[tex]v_f = 72 km/h=20 m/s[/tex]

[tex]v_i = 54 km/h = 15 m/s[/tex]

[tex]a = 2 m/s^2[/tex]

2. Find time first.

[tex]t = \frac{v_f - v_i}{a}[/tex]

[tex]t = \frac{20 - 15}{2}[/tex]

[tex]t= 2.5[/tex] seconds

2. Then find the distance.

[tex]s = v_i t \frac{1}{2} a t^2[/tex]

[tex]s = (15)(2.5)\frac{1}{2} (2)(2.5)^2[/tex]

[tex]s = 234.4[/tex] m

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

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​

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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