Question

4) A racing car undergoing constant acceleration covers 140 m in 3.6 s. (a) If it’s moving at 53 m/s at the end of this interval, what was its speed at the beginning of the interval? (b) How far did it travel from rest to the end of the 140-m distance

Answer 1

To find the initial velocity of the racing car, we can use the equation v_f = v_i + at. Given that the final velocity is 53 m/s and the time is 3.6 seconds, we can rearrange the equation to solve for the initial velocity. Additionally, we can calculate the distance traveled by the car using the equation d = v_i t + 0.5at^2.

Step-by-step explanation:

To find the initial velocity of the racing car, we can use the equation:

vf = vi + at

where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.

Given that the final velocity is 53 m/s, the acceleration is unknown, and the time is 3.6 seconds, we can rearrange the equation to solve for the initial velocity:

vi = vf - at

Substituting the known values:

vi = 53 m/s - (a)(3.6 s)

We also know that the car covers a distance of 140 meters, which can be calculated using the equation:

d = vit + 0.5at2

Plugging in the values:

140 m = (vi)(3.6 s) + 0.5(a)(3.6 s)2

By solving these two equations simultaneously, we can find the initial velocity and acceleration of the racing car.

Answer 2

given,

distance = 140 m

time, t = 3.6 s

moving speed = 53 m/s

a) distance = (average velocity) x time

`D = (v_0 + v_1)/(2)* t`

`140 = (v_0 + 53)/(2)* 3.6`

v₀ + 53 = 77.78

v₀ = 24.78 m/s or 25 m/s

b)
`a = (v-u)/(t)`

`a = (53-25)/(3.6)`

a = 7.8 m/s²

using equation of motion

v₀² = v₁² + 2 a s

53² = 0²+ 2 x 7.8 x s

s = 180 m