Question

A particle starting from rest moves with a constant acceleration if it takes 5 second to reach the speed 18 km per hour find average velocity during this period and the distance travelled by the particle during this period.

Answer 1

Explanation:

initial velocity ,u = 0 m/s

time =5 second

final velocity, v = 18km /hr

$= 18 \times \frac{5}{18}m \: {s}^{ - 1} = 5m \: {s}^{ - 1}$

acceleration,a=(v -u)/t

=

$\frac{5-0}{5} = 1m \: {s}^{ - 2}$

Distance

according to 2nd equation of motion

$s = ut + \frac{1}{2} a {t}^{2} \\ s = 0 \times 5 + \frac{1}{2} \times 1 \times 5\times 5 \\ s = 12.5m$

average speed

$= \frac{total \: distance \: travelled \: }{total \: time \: taken \: } \\ = \frac{12.5}{5} = 2.5m \: {s}^{ - 2}$

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