Question

A car is moving with a uniform velocity of 10m/s. the driver of the car decides to
overtake the bus moving ahead of the car. So the driver of the car accelerates at 1m/s2 for 10
sec. Find the velocity of the car at the end of 10 sec. also find the distance traveled by the
car while accelerating
pls answer with explaination
and be fast

Answer 1

we have initial velocity which is 10m/sec and we have time 10sec and acceleration 1m/sec2
we will apply equation to find acceleration so
a=v-u÷t
a=v-10÷10
1×10=v-10
10+10=v
v=20m/sec
then we hAve to find distance
we will apply first equation of motion
s=ut+1/2at2
so
s=10×10+1/2×1×{10}2
s=100+50
s=150m
hope it was helpful
please mark it as brainliest answer

Answer 2

Hey sissy !!!!

•°• Here's your answer •°•

Given =

Initial Velocity = u = 10m/s

Acceleration = a = 1m/s^2

Time = t = 10sec

Final Velocity = v = ??

Distance = s = ??

The three equations of motion are =

$v = u + at \\ \\ s = ut + \frac{1}{2} a {t}^{2} \\ \\ {v}^{2} = {u}^{2} + 2as$

Let's find v by first equation of motion

$v = u + at \\ \\ v = 10 + 1 \times 10 \\ \\ v = 10 + 10 \\ \\ v = 20$
So v = 20m/s

Now, let's find s by 2nd equation of motion

$s = ut + \frac{1}{2} a {t}^{2} \\ \\ s = 10 \times 10 + \frac{1}{2} \times 1 \times {10}^{2} \\ \\ s = 100 + 50 \\ \\ s = 150$

Let's find s by 3rd equation of motion

${v}^{2} = {u}^{2} + 2as \\ \\ {20}^{2} = {10}^{2} + 2 \times 1 \times s \\ \\ 400 = 100 + 2s \\ \\ 400 - 100 = 2s \\ \\ \frac{300}{2} = s \\ \\ 150 = s$

So your answers are as follows

Final Velocity = v = 20m/s

Distance = s = 150m

Hope it satisfies you ☆▪☆

Thanks ^_^

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