Question

A sport car moving at constant speed travels 110m in 5.0 s. if it then brakes and comes to a stop in 4.0 s, what is the magnitude of its acceleration in m/s2, and in g’s (g=9.80 m/s2)? 2

Answer 1

The sports car decelerates at a rate of -5.5 m/s² or -0.56 g's (where g = 9.8 m/s²) when it comes to a stop in 4.0 seconds from a speed of 22m/s.

Step-by-step explanation:

The sports car is initially moving at a constant speed, which means the acceleration is zero during that phase. However, when the car applies the brakes and comes to a stop, it undergoes a negative acceleration or deceleration. To find this deceleration, we need to calculate the change in velocity over the time taken to stop.

Firstly, we find the initial speed of the car: The car covers 110m in 5.0s, which gives us a speed (v) using the formula v = d/t, where d is distance and t is time. Hence, v = 110m / 5.0s = 22m/s. Next, we use the deceleration formula a = Δv/Δt, where Δv is the change in velocity and Δt is the change in time. The change in velocity is from 22m/s to 0m/s, so Δv = 0m/s - 22m/s = -22m/s, and the time taken to stop is Δt = 4.0s.

The deceleration a = -22m/s / 4.0s = -5.5 m/s². To convert this to g's, remember that 1g = 9.80m/s², so a(g's) = -5.5m/s² / 9.80m/s² = -0.56 g's.

Answer 2

d = 110 m, t = 5 s
v o = 110 cs : 5 m = 22 m/s
-------------------------------------
v = v o - a t
v = 0 m/s, v o = 22 m/s, t = 4 s
0 = 22 - 4 a
4 a = 22
a = 22 : 4
a = 5.5 m/s²
g = 9.80 m/s²
9.80 : 5.5 = 0.56
Answer:
The magnitude of its acceleration is 5.5 m/s or 0.56 g.