Question

A car accelerates uniformly from 18km h-¹ to 36 km h-¹ in 5 s. Calculate i) the acceleration and ii) the distance covered by the car in that time.​

Answer 1

initial velocity = 18 km/h = 5 m/s

final velocity = 36 km/h = 10 m/s

time = 5 s

(I)

$acceleration = \frac{(v - u)}{t}$

a = 10 - 5/5

a = 5/5

a = 1

car accelerated 1 m/s²

(ii)

$distance = ut + \frac{1}{2} a{t}^{2}$

s = (5×5) + 1/2 × 5 × 25

s = 25 + 62.5

s = 87.5

the distance covered by the car is 87.5 m

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Answer 2

Answer:

☆ To find:—

▪︎Acceleration =?

▪︎Distance =?

☆ Formula applied:—

$a = \frac{v - u}{t}$

and

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

☆Solution :—

Given that,

u = 18 m/s

= 18×1000m/60×60s

= 5 m/s

v = 36 km/h

= 36×1000m/60×60s

= 10 m/s

t = 5 s .

(i) From equation (8.5) we have,

$a = \frac{v - u}{t}$

$= \frac{10m {s }^{ - 1} - 5m {s}^{ - 1} }{5s}$

$= 1m {s}^{ - 2}$

(ii)From equation (8.6) we have,

s = ut + ½ at²

= 5 m/s × 5 s + ½ × 1 m/s × (5 s)²

= 25 m + 12.5 m

= 37.5 m

Therefore, the acceleration of the car is 1 m/s and the distance covered is 37.5 m.