Question

A car starting from rest accelerates uniformly to
acquire a speed 20 km h-1 in 30 min. The distance
travelled by car in this time interval will be :​

Answer 1

Answer:

distance covered = 4860 meter

Step by step explanations :

given that,

A car starting from rest

accelerates uniformly to acquire

a speed 20 km/h in 30 min

here,

initial velocity of the car = 0 m/s

[because car was initially at rest]

final velocity of the car = 20 km/h

= 20 × 1000 m/3600 s

= 50/9 m/s

time taken to accelerate = 30 minutes,

= 30 × 60 s

= 1800 s

let the distance covered by the car during acceleration be s

now,

we have,

initial velocity(u) = 0 m/s

final velocity(v) = 50/9 m/s

time taken(t) = 1800 seconds

by the equation of motion,

v = u + at

putting the values,

50/9 = 0 + a(1800)

1800a = 50/9

a = 50/9 × 1/1800

a = 0.003 m/s²

also,

S = ut + ½ at²

again putting the values,

S = 0(1800) + ½ × 0.003 × 1800 × 1800

S = 4860 m

so,

distance covered by the car during acceleration

= 4860 meter

Answer 2

Answer:

4860 m

Explanation:

Given

a car starting from rest (u=0), accelerates uniformly to acquire the speed 20 km/h in 30 mins

intial velocity (u) = 0

Final Velocity (v) = 20 × $\dfrac{5}{18}$ = $\dfrac{50}{9}$ m/s

time (t) = 30 × 60 = 1800 sec

To Find

The distance travelled by the car (S)

Solution

From the 1st equation of motion

$\boxed{v=u+at}$

$\dfrac{50}{9}=0+a(1800)$

$\dfrac{50}{9}=1800a$

$a = \dfrac{50}{9 \times 1800}$

$a = 0.003 m/{s}^{2}$

From the 2nd equation of motion

$\boxed{S=ut+\dfrac{1}{2}a{t}^{2}}$

$S=(0)t+\dfrac{1}{2}(0.003)({1800})^{2}$

$S=\dfrac{1}{2} \times \dfrac{3}{1000} (1800 \times 1800)$

$S= \dfrac{3}{2000} (1800 \times 1800)$

$S=3(180)(9)$

$S=4860 m$

$S=4.86 km$