A car accelerates from 14.0m/s to 21m/s in 6.0s. What was its acceleration? How far did it travel in this time? Assume constant acceleration.

Acceleration (a) = 1.167 m/s²

Distance travel = 105 m (Approx)

Explanation:

Given:

Initial speed (u) = 14 m/s

Final speed (v) = 21 m/s

Time taken (t) = 6 sec

Find:

Acceleration

Distance travel

Computation:

v = u + at

21 = 14 + a(6)

7 = 6a

Acceleration (a) = 1.167 m/s²

S = ut + (1/2)(a)(t²)

S = (14)(6) + (1/2)(1.167)(6²)

S = 84 + 21

S = 105 m (Approx)

Distance travel = 105 m (Approx)

ι вєℓιєνє ιт ιѕ тнє fιяѕт ¢нσι¢є(α.)

нє ѕнσυℓ∂ ѕмιℓє ρσℓιтℓу αи∂ ѕρєαк тσ тнє ρєяѕσи ωнσ ιѕ ιитєяяυριиg αfтєя тнє ∂ιѕ¢υѕѕισи

нσρє ι ¢συℓ∂ нєℓρ.