Question

A racing car’s velocity is increased from 44 m/s to 66 m/s in 11 seconds. What is the acceleration? and what is the displacement?

Answer 1

The acceleration of the racing car is 2 m/s² and the displacement over the 11 seconds is 605 m.

Step-by-step explanation:

The given physics problem involves calculating the acceleration and the displacement of a racing car that has increased its velocity from 44 m/s to 66 m/s in a time interval of 11 seconds.

To find the acceleration, we use the formula for average acceleration, which is the change in velocity divided by the time taken. The change in velocity (Δv) is 66 m/s - 44 m/s = 22 m/s. So, the acceleration (a) is 22 m/s divided by 11 s, which equals 2 m/s².

For the displacement, we can use the kinematic equation s = ut + 0.5at², where u is the initial velocity, a is the acceleration, and t is the time. With u = 44 m/s, a = 2 m/s², and t = 11 s, we get the displacement s = 44 m/s * 11 s + 0.5 * 2 m/s² * (11 s)², resulting in 484 m + 121 m, or a total displacement of 605 m.

Answer 2

Step-by-step explanation:

Acceleration is defined as the change in velocity per unit of time

Acceleration (a) = ΔV/t

V = Velocity t = time

ΔV = V₂ - V₁ t = 11s

V₁ = 44m/s

V₂ = 66m/s

ΔV = 66 - 44

= 22m/s

Acceleration (a) = 22/11

= 2m/s²

Displacement (d): Displacement equals the original velocity multiplied by time plus one half the acceleration multiplied by the square of time.

d = v₀ + 1/2*at²

v₀ = 44m/s

a = 2m/s

t = 11s

d = 44 + 1/2*(2 x 11²)

= 44 + 121

d = 165m