Question

A car starts from rest and travels for6.0 s with a uniform acceleration of+1.6m/s². The driver then applies the brakes, causing auniform acceleration of-1.0 m/s². If the brakes are applied for2.0 s, determine each of thefollowing.(a) How fast is the car going at the end of thebraking period?
1 m/s
(b) How far has it gone?
2 m

Answer 1

for the first 6 s,
distance = (1/2)at² = (.5)(1.6)(36) m
speed at the end of 6 s = at = (1.6)(6) m/s = 9.6 m/s
for the next 2 s,
distance = ut - (1/2)at² = (9.6)(2) - (.5)(1)(4) m
(a)
speed at the end = 9.6 - (1)(2) = 7.6 m/s
(b)
total distance = (.5)(1.6)(36) m + (9.6)(2) - (.5)(1)(4) m
compute the numbers.
hope this helps!

Answer 2

GIVEN DATA :initial velocity = v = 0 m/stime taken= t =6 secondsuniform acceleration = a = +1.6 m/s²uniform deceleration = -1.0 m/s²brakes are applied for 2.0 sec.final velocity = v' = ?distance covered = s = ?SOLUTION :for time t = 6 sec ,velocity of the car = 0 +(1.6) (6) = 9.6 m/s.hence when brakes are applied then velocity ofcar =9.6 m/s.using equation v' = v + atso required final velocity of car = v'= 9.6+(-1) (2) =7.6 m/s.using s = vt + 1/2 at²while distance covered = s =(9.6) (2) +(0.5) (-1) (2)² = 19.2 - 2 = 17.6 meter