Question

A racing car moving with constant acceleration covers two successive kilometers in 30s and 20s respectively. Find the acceleration of the car. Ans= 2/3 m/s2 PLEASE HELP! :O

Answer 1

Answer Let initial velocity= u

accl^n=a

A/Q_

S= ut + 1/2 at^2

1000=u×30 + 1/2 × a×30×30

1000=30u + 450a _____(1)

Velocity after 1km= u + at =u+30a

S= ut +1/2at^2

1000= (u +30a)×20 +0.5a×(20)^2

1000=20u +600a+200a

1000=20u+800a

u =35a______(2)

Putting the value of (2) in (1)

1000=30u +450a

1000=30×35a+450a

1000=1050a+450a

1000=1500a

a =1000/1500

a=2/3 m/s^2

Hope u understand this.....

Explanation:

Answer 2

Answer:

Acceleration of the car is $\frac{2}{3} \mathrm{~ms}^{-2}$

Explanation:

Given,

Acceleration covers two successive kilometers in the 30s and 20s.

To find,

The acceleration of the car.

• Acceleration exists as a vector quantity that is defined as the rate at which an object alters its velocity. An object is accelerating if it is altering its velocity.
• Acceleration is the rate at which velocity differs with time, in phrases of both speed and direction.

$v_{1}^{2}-u_{1}^{2}=2 a s . \ldots \ldots \ldots .(1)$

$v_{1}=u_{1}+a(30) ;-30$seconds travel

$v_{2}^{2}-u_{2}^{2}=2 a s . \ldots \ldots \ldots(2)$

$v_{2}=u_{2}+a(20) ;-20$seconds travel

Here $v_{1}=u_{2}$

therefore,

Adding $(1),(2)$

$v_{1}^{2}-u_{1}^{2}=4 a s$

$\left(v_{2}-u_{1}\right)\left(v_{2}+u_{1}\right)=4 a s$

Then,

$v_{2}=u_{2}+a(20)$

$u_{2}=v_{1}=u_{1}+a(30)$

$\Rightarrow v_{2}=u_{1}+a(30)+a(20)$

We get,

$=u_{1}+50 a$

Substituting in (3)

$&\left(u_{1}+50 a-u_{1}\right)\left(u_{1}+u_{1}+50 a\right)=4 a s$

$&50 a\left(2 u_{1}+50 a\right)=4 a s$

$&\Rightarrow 50\left(2 u_{1}+50 a\right)=4 s$

We get,

$&s=1000 m$

$&\Rightarrow 50\left(2 u_{1}+50 a\right)=4 \times 1000$

$&\Rightarrow 2 u_{1}+50 a=\frac{4000}{50}=80$

$&\Rightarrow 2 u_{1}+50 a=80$

Dividing by 2 on both sides

$\mathrm{u}_{1}+25 \mathrm{a}=40 \ldots \ldots \ldots \ldots \ldots(4)$

$&v_{1}^{2}-u_{1}^{2}=2 a s$

$&\left(v_{1}-u_{1}\right)\left(v_{1}+u_{1}\right)=2 a s$

$&v_{1}=u_{1}+30 a, \quad s=1000$

$&\left(u_{1}+30 a-u_{1}\right)\left(u_{1}+30 a+u_{1}\right)$

$&=2 a \times 1000$

$&30 a\left(2 u_{1}+30 a\right)=2000 a$

$&2 u_{1}+30 a=\frac{200}{3}$

$&u_{1}+15 a=\frac{100}{3}$

Subtracting (4)-(5)

$u_{1} &+25 a-\left(u_{1}+15 a\right)$

$&=40-\frac{100}{3}$

$10 a &=\frac{2.0}{3}$

$a &=\frac{2}{3}$

acceleration of the car is $\frac{2}{3} \mathrm{~ms}^{-2}$

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