Question

A car acquires a velocity of 72km/h in 10s. starting from rest find.
1. the occuleration
2. The average velocity
3. Distance travellled in this time

Answer 1

Given initial velocity u=0 final velo v=72km/h = 72*1000/3600=20m/s time t=10s acceleration a=(v-u)/t=(20-0)/10=2m/s sq. average velo=( u+v)/2= 10m/s dist s=ut+(1/2)at^2=(1/2)(2)(10)(10)=100m

Answer 2

Answer- The above question is from the chapter 'Kinematics'.

Some important terms and formulae:-

1. Velocity- It is the displacement per unit time.

S.I. Unit of Velocity- m/s

It is a vector quantity as it possesses magnitude and direction.

2. Acceleration- It is the rate of change of velocity.

S.I. Unit of Acceleration- m/s²

It is also a vector quantity.

Negative acceleration is called retardation.

3. Distance- It is the path length transversed by an object.

S.I. Unit of Distance- m

It is a scalar quantity.

4. Displacement- It is the shortest distance between the initial and final point.

S.I. Unit of Displacement- m

It is a vector quantity.

5. Equations for uniformly accelerated motion-

Let u = Initial velocity of a particle

v = Final velocity of a particle

t = Time taken

s = Distance travelled in the given time

a = Acceleration

1) v = u + at

2) s = $\frac{1}{2}$ at² + ut

3) v² - u² = 2as

Given question: A car acquires a velocity of 72km/h in 10 s starting from rest. Find:

1. The acceleration

2. The average velocity

3. Distance travelled in this time

Answer: Initial velocity (u) = 0 m/s

Final velocity (v) = 72 km/h = 20 m/s

Time (t) = 10 s

1. Using first equation of motion,

v = u + at

20 = 0 + 10a

10a = 20

∴ a = 2 m/s²

2. Since the acceleration is constant,

average velocity = $\frac{u \: + \: v}{2}$

                             = $\frac{0 \: + \: 20}{2}$

∴ average velocity = 10 m/s

3. Let s be the distance travelled.

s = $\frac{1}{2}$ at² + ut

s = $\frac{1}{2}$ 2 × 10² + 0 × 10

s = 100 + 0

∴ s = 100 m