Question

A car start from rest and acquire a velocity of 54 km/h in 2 sec. Find
(i) the acceleration 
(ii) distance travelled by car assume motion of car is uniform?​

Answer 1

speed of car in m/s=54× 5/18

=15

acceleration= 15/2

=7.5

if the speed of the car is uniform at 15 m/s then the distance travelled would be 30m

hope it helps please mark brainliest

Answer 2

Explanation:

From the question,

Final velocity,v:54km/h

Time taken,t:2sec

The car is starting from rest,so the initial velocity of the car is zero

•u:0m/s

As the other quantities are in mks system,converting 54km/h into m/s scale,

$\mathsf{54 \times \frac{5}{18} } \\ \\ = \mathsf{3 \times 5} \\ \\ = \mathsf{15ms {}^{ - 1} }$

Thus,final velocity is 20m/s

1. Acceleration of the car:

Acceleration is the change of velocity of an object per unit time.

$\mathsf{a = \frac{v - u}{t} } \\ \\ \implies \: \mathsf{a = \frac{15}{2} } \\ \\ \implies \: \boxed {\mathsf{a = 7.5ms {}^{ - 2}}}$

Acceleration of the car is 7.5m/s²

Now,

•Distance travelled by the car:

Using the formula,

$\mathsf{v {}^{2} - u {}^{2} = 2as }$

Here,

s is the distance travelled,a is the acceleration,u is the initial velocity and v is the final velocity.

Putting the values,

$\mathsf{15 {}^{2} = 2(7.5)s } \\ \\ \implies \: \mathsf{s = \frac{225}{15} } \\ \\ \implies \: \mathsf{s = 15m}$

The distance travelled by the car is 15m