Question

Amit accelerates her scooty uniformly from 18 km/h to 63 km/h in 5s. Calculate the (a) acceleration (b) distance covered by him in that time.(3)​

Answer 1

Answer:

acceleration =  9 kmh⁻²

distance covered by him in that time = 212.5 km

Step-by-step explanation:

Given:

u (initial velocity) = 18 km/h

v (final velocity)= 63 km/h

t (time) = 5s

(i) acceleration

a (acceleration) = ?

$a = \frac{v-u}{t} \\\\a =\frac{ 63 - 18}{5}\\\\a = \frac{45}{5} \\\\a = 9$

∴ acceleration = 9 km/h² = 9 kmh⁻²

(ii) distance covered by him in that time

s (distance) = ?

$s = ut +\frac{1}{2} at^{2} \\\\substituting\ all\ the\ values\\\\s = (18)(5) +\frac{1}{2} (9)(5)^{2} \\\\s = (90) +\frac{1}{2} (9)(25)\\\\ s = 90 +\frac{1}{2} (225)\\\\ s = 90 +112.5\\\\s=212.5$

∴distance = 212.5 km

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