Question

A car travelling at 9m/s accelerates and attains speed of 27m/s in 6s. Find the acceleration and distance travelled during this time

Answer 1

Answer:

4 km

Explanation:

27/9×3

=9/3×2

=2×2

=4

Answer 2

A car travelling at 9 m/s accelerates and attain a speed of 27 m/s in 5 s, what distance does it cover?

V^2=u^2 + 2*a*s

V is final velocity.

u is initial velocity.

a is acceleration.

s is distance.

$27^2=9^2+ 2 (\frac{27–9}{5s})$

729=81 + 2 (7.2) s

648=7.2 s

90 m =s

List the known and unknown variables.

u = initial velocity = 9 m/s

v = final velocity = 27 m/s

∆t = time interval = 5 s

d = displacement = ? m

Use the following kinematic equation to calculate the displacement of the car.

d = (v - u)

d = $\frac{v - u}{2 × t}$

d = $\frac{9 + 27}{2 × 5 }$

d = $\frac{36} {2 × 5}$ = 90 m

$\rule{770mm}{3pt}$