Question

A car traveling 77 km/h slows down at a constant 0.48 m/s2 just by "letting up on the gas." A) Calculate the distance the car coasts before it stops. B)Calculate the time it takes to stop. C)Calculate the distance it travels during the first second. D)Calculate the distance it travels during the fifth second.

Answer 1

Answer:

(a) 477 m

(b) 44.6 s

(c) 21.16 m

(d) 19.24 m

Explanation:

initial speed, u = 77km/h = 21.4 m/s

acceleration, a = - 0.48 m/s2

final speed v =  0  

(a) let the stopping distance is s.

Use third equation of motion

$v^2 = u^2 + 2 a s\\\\0 = 21.4^2 - 2 \times 0.48\times s\\\\s = 477 m$

(b) Let t is time.

Use first equation of motion

v = u + at

0 = 21.4 - 0.48 t

t = 44.6 s

(c) Let the distance is s in first second.

Use second equation of motion

$s = u t + 0.5 at^2\\\\s = 21.4\times 1 - 0.5\times 0.48\times 1\\\\s = 21.4 - 0.24 = 21.16 m$

(d) distance traveled in 5 th second is given by

$s = u + 0.5 a (2 n - 1) \\\\s = 21.4 - 0.5\times 0.48 \times (2\times 5 -1)\\\\s= 21.4 - 2.16 = 19.24 m$