Question

A scooter of mass 110 kg travelling with speedof 36 km/hr comes to rest in 5 s after applying
breaks. What is its deceleration? How much
force has acted upon the scooter? [3 M]​

Answer 1

Answer:

$\frac{-2}{5} m/s^{2}$

$F= -44 N$

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Explanation:

Here,

U= 36 km/ hr

When changed to m/ sec it is

36 x 5/18

= 2 x5

=10 m/sec

mass= 110kg  

t=5sec

and as given it stops,

so Vfinal= 0m/s

deceleration = Δv/Δt = v-u/$t^{2}$

=    $\frac{0-10}{5^{2} } m/s^{2}$

=    $\frac{-10}{25} m/s^{2}$

=   $\frac{-2}{5} m/s^{2}$

we know,

$F=ma$

F= 110 × -2/5

F = 22× -2

$F=-44$ N