Question

A car is accelerated uniformly from 36 km/h to 54km/h in 10 seconds.find: i) the acceleration and ii) the distance travelled by car during this interval of time.

Answer 1

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HERE IS YOUR ANSWER :-

GIVEN :

INITIAL VELOCITY (u) = 36 Km/h

INITIAL VELOCITY (u) = 10 m/s

FINAL VELOCITY (v) = 54 Km/h

FINAL VELOCITY (v) = 15 m/s

TIME = 10 SECONDS

I) ACCELERATION = v - u/t

ACCELERATION = 15 - 10/10

ACCELERATION = 5/10

ACCELERATION = 0.5 m/s^2

II) DISTANCE (s) = ut + 1/2 at^2

DISTANCE (s) = 10 × 10 + 1/2 × 0.5 × 10^2

DISTANCE (s) = 100 + 25

DISTANCE (s) = 125 m
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Answer 2

Hello!

A car is accelerated uniformly from 36 km/h to 54km/h in 10 seconds.find: i) the acceleration and ii) the distance travelled by car during this interval of time.

i) the acceleration

We have the following data:

V (final velocity) = 54 km/h (in m/s) = 54 / 3.6 = 15 m/s

Vo (initial velocity) = 36 km/h (in /m/s) = 36 / 3.6 = 10 m/s

ΔV  (speed interval)  = V - Vo → ΔV  = 15 - 10 → ΔV  = 5 m/s

ΔT (time interval) = 10 s

a (average acceleration) = ? (in m/s²)

Formula:

$\boxed{a = \dfrac{\Delta{V}}{\Delta{T^}}}$

Solving:  

$a = \dfrac{\Delta{V}}{\Delta{T^}}$

$a = \dfrac{5\:\dfrac{m}{s}}{10\:s}$

$\boxed{\boxed{a = 0.5\:m/s^2}}\longleftarrow(acceleration)\end{array}}\qquad\checkmark$

ii) the distance travelled by car during this interval of time.

We have the following data:

Vi (initial velocity) = 36 km/h (in /m/s) = 36 / 3.6 = 10 m/s

t (time) = 10 s

a (average acceleration) = 0.5 m/s²

d (distance travelled) = ? (in m)

By the formula of the space of the Uniformly Varied Movement, it is:

$\boxed{d = v_i * t + \dfrac{a*t^{2}}{2}}$

Solving:

$d = v_i * t + \dfrac{a*t^{2}}{2}$

$d = 10 * 10 + \dfrac{0.5*(10)^{2}}{2}$

$d = 100 + \dfrac{0.5*100}{2}$

$d = 100 + \dfrac{50}{2}$

$d = 100 + 25$

$\boxed{\boxed{d = 125\:m}}\Longleftarrow(distance\:travelled)\end{array}}\qquad\checkmark$

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