Question

QUESTION;
A car moving with constant acceleration covers the distance between two points 180 m apart in 6 sec.

Its speed as it passes the second point is 45 m/s. Its acceleration . [ ]

a) – 5 m/s² b) – 15 m/s²

c) 25 m/s² d) 5 m/s²

By Yagnasri,✅✅✔️☑️
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Answer 1

Answer:

your answer is 15

Explanation:

because the car is moving from point A to b is 30m/s constantly.

then from b to c it had speed of 45 m/s.

It increase by 15m/s.

180=u(6)+1/2a (6)square

180=6u+18a

30=u+3a......(eq 1)

45=u+at

45=u+a (6)....(eq 2 ) because time is 6 sec from point A to b.

then( eq 2) -(eq 1)

(45-30=15)

15=3a then a=15÷3=5 then a=5

then ,

u=15 m/s.

Answer 2

Answer:-

➣

• Given:-

$\implies{Speed \: of \:car}= {\bf{45 m/s}}$

$\implies{Distance}= {\bf{180m}}$

$\implies{Time \: taken}={\bf{6 \: seconds}}$

• To Find:-

Acceleration of car =$\bf{?}$

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• Solution:-

Distance between two points (s) = $\bf{180 m}$

Time taken to cover the distance (t) = $\bf{6 s}$

Final velocity (v) = $\bf{45 m/s}$

Therefore, Finding the initial velocity and acceleration:-

• We know:-

The 1st equation of motion is:-

$\boxed{\bf{v \: = \: u \: + \: at}}$

Hence,

$\longrightarrow{45 \: = \: u \: + 6a}$

$\longrightarrow{u = 45 - 6a}\: \: \: \longrightarrow {\bf(eqn.1)}$

Using the 2nd equation of motion:-

• We have:-

$\boxed{\bf{s = ut + \frac{1}{2} \: a {t}^{2}}} \: \: \: \longrightarrow {\bf(eqn.2)}$

• Substituting (eqn.1) in (eqn.2):-

$180 = (45 - 6a)6 + ( \frac{1}{2} a \times 36)$

$180 = 270 - 36a + \frac{36a}{2}$

$180 = 270 - 36a(1 - \frac{1}{2})$

$180 = 270 - 36( \frac{1}{2} )$

$180 = 270 - 18a$

$18a = 270 - 180$

$18a = 90$

$a = \frac{90}{18}$

$a = {\cancel{\frac{90}{18}}}$

$\implies \bf{5 \: m {s}^{ - 2}}$

Putting, the value of acceleration [a] in equation (1):-

We have:-

$u = 45 - (6 \times 5)$

$u = 45 - 30$

$\bf{u = 15 \: m {s}^{ - 1}}$

Therefore,

• $\large \bf{Initial \: velocity[u] = 15 m/s^{-1}}$

• $\large \bf{acceleration[a]= 5m/s^{-2}}$

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