Question

17. A car accelerates uniformly from 18 kmh-' to 36 kmh ' in 2 seconds.
Calculate.
(a) Acceleration
(b) Distance covered by the car in that time.
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Answer 1

Explanation:

By substituting the values in the equation we will get acceleration as 2.5 m/s2 and distance as 15m

Answer 2

Explanation:

(a) Acceleration = (v - u) / t

a = (36 - 18 ) / 2

a = 18/2

a = 9 m/s^2

(b) Using

${v}^{2} = \: {u}^{2} + 2 \times a \times s$

Here, v is final velocity.

u is initial velocity.

a is acceleration.

s is displacement covered.

So, We get

${36}^{2} = {18}^{2} + 2 \times 9 \times s$

$1296 - 324 \: = \: 18s$

$972 \div 18 = s$

Hence,

$s = 54 \: m$

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