Physics: 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

a car moving with a velocity of 54km/h accelerate

Ree0628

20.09.2021

ANSWER: 234.4 m and 2.5 seconds

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.

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

$t = \frac{20 - 15}{2}$

t= 2.5 seconds

2. Then find the distance.

$s = v_i t \frac{1}{2} a t^2$

$s = (15)(2.5)\frac{1}{2} (2)(2.5)^2$

s = 234.4 m

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

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asra44

13.12.2022

P = 782 N

g = 9.81 m/s^2

So:

m = P/g = 782N/9.81 m/s^2 = 79,7 Kg

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