Question

A motor bike running at 90 kmh-1 is slowed down at 18kmh-1 in 2.5m. calculate: a. Acceleration. b. Distance covered in time it’s slow down.

Answer 1

v = u + at
90×5/18 = 18×5/18 + a × 2.5×60
25 = 5 + 15t
t = 20/15 = 4/3 m/s^2
s = ut + 1/2at^2
s = 25×15 + 1/2 ×4/3 × 15×15

Answer 2

Explanation :

(1) Initial velocity of motor bike, $u=90\ km/h=25\ m/s$

Final velocity of motor bike, $v=18\ km/h=5\ m/s$

Time taken, $t=2.5\ m=150\ s$

Acceleration, $a=\dfrac{v-u}{t}$

$a=\dfrac{5\ m/s-25\ m/s}{150\ s}$

$a=-0.13\ m/s^2$

-ve sign shows that the motorbike is decelerating.

(2) Using third equation of motion :

$(5\ m/s)^2-(25\ m/s)^2=2\times-0.13\ m/s^2\times s$

s is the distance covered.

$s=2307\ m$

Distance covered is 2307 m.