Question

If a car at rest accelerates uniformly to a speed of 144 km/hr in 20 sec,it covers a distance of,. a.1440 cm. b.2980 cm. c.20 m. d.400 m.​

Answer 1

Given :

• initial velocity = 0 m/s
• final velocity = 144 km / hr
• time taken = 20 sec.

Unit conversion :

( In order to convert km /hr into m/s multiply by 5 / 18 )

final velocity

$v = 144 \times \frac{5}{18} = 40 \frac{m}{s}$

Solution :

First we need to find the acceleration of the car

By using the first kinematic equation ,

$\displaystyle \star\boxed{ \blue{v = u + at}}$

here ,

• v = final velocity
• u = initial velocity
• a = acceleration
• t = time

Substituting the given values in the above equation we get ,

$\implies{v = at}$

$\implies{a = \frac{v}{t} }$

$\implies{a = \frac{40}{20} }$

$\boxed{ \green{a = 2 \frac{m}{ {s}^{2} } }}$

Now , in order to find distance we'll use the second kinematic equation ,

$\star\boxed{ \blue{s = ut + \frac{1}{2} a {t}^{2} }}$

here ,

• s = distance

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

$\implies{s = \frac{1}{2} \times 2 \times 400}$

$\boxed{ \red {s = 400m}}$

Answer 2

Answer:

(d) 400 metres

Explanation:

Given:

• Initial velocity = u = 0 m/s
• Final velocity = v = 144 km/hr = 144×5/18 = 40 m/s
• Time taken = t = 20 seconds

To find:

• Distance covered by the car with the above etails

First using the first equation of motion: v=u+at

40=0+a×20

40=20a

a = 40/20

a = 2 m/s²

Using the second equation of motion:

s=ut+1/2×at²

s=0×20+1/2×2×20²

s=0+1×400

s=400 metres

The distance covered by the car is equal to 400 metres

Hence option (d) is the right answer