Question

You are driving in such a way that the car is accelerating at a constant rate in the positive direction. When you pass the first sign, you are traveling at 4 m/s. When you pass the second sign 50 m down the road, you note that the seconds indicator of your clock reads 45 seconds. You also note that your velocity is now 9 m/s.
Required:
a. What is your acceleration?
b. What was the clock’s seconds indicator reading when you passed the first sign?

Answer 1

Answer:

Explanation:

a)

v² = u² + 2 a s

v = 9 m/s

u = 4 m/s

s = 50 m

9² = 4² + 2 x a x 50

a = 0.65 m /s²

Acceleration is 0.65 m /s²

b )  

time elapsed before velocity changed from 4 m/s to 9 m/s with acceleration of .65 m /s ²

(v - u ) / t = a

(v - u ) / a = t

(9 - 4 ) / .65 = t

t = 7.7

time when passing the first sign will be 7.7 s earlier .

Reading of time indicator = 45 - 7.7

= 37.3 seconds.

Answer 2

Answer:

(a) 0.45 m/s^2

(b) 33.9 s

Explanation:

initial velocity, u = 4 m/s

final velocity, v = 9 m/s

distance, s = 50 m

(a) Let the acceleration is a.

Use third equation of motion

$v^2 = u^2 + 2 as \\\\9^2 = 4^2 + 2\times a\times 50\\\\a = 0.45 m/s^2$

(b) Let the time is t.

Use first equation of motion

v = u + at

9 = 4 + 0.45 x t

t = 11.1 s

So, the initial time, t' = 45 - 11.1 = 33.9 s