A boy riding a bicycle applies brake and slows down at a rate of 5 m/s square for a distance of 8 m. If the speed of bike was 12 m/s when he applied the brakes,
a. what is his final velocity?
b. At this deceleration how long would it take him to stop?
Answers
Answer:
a) Final velocity of boy after traveling 8 metres at given deceleration will be 8 ms⁻¹
b) Time taken by boy to come to rest will be 2.4 seconds.
Explanation:
acceleration of bicycle , a = - 5 ms⁻² (negative acceleration shows deceleration)
distance travelled after applying breaks, s = 8 m
initial velocity of boy (when he applied breaks), u = 12 ms⁻¹
we need to find the final velocity (v) of boy after decelerating for 8 metres of distance
Using third equation of motion
→ 2 a s = v² - u²
→ 2 (-5) (8) = v² - (12)²
→ -80 = v² - 144
→ v² = 144 - 80
→ v = 8 ms⁻¹
therefore,
final velocity of boy after decelerating for 8 metres of distance will be 8 ms⁻¹.
Now,
we need to find how long would it take him to stop.
therefore,
final velocity , v = 0
initial velocity of boy , u = 12 ms⁻¹
acceleration of boy, a = -5 ms⁻²
Using first equation of motion
→ v = u + a t
→ 0 = 12 + (-5) t
→ -12 = -5 t
→ t = 12 / 5
→ t = 2.4 s
therefore,
Time taken by boy to stop is 2.4 seconds.