Question

A car accelerates uniformly from rest at a rate of 10ms-² on a straight level road. calculate the displacement of the car. 1) after 5 seconds 2)in the 5th second of motion​

Answer 1

Answer:

HEY BUDDY! HERE IS YOUR ANSWER.

Explanation:

IN Ist CASE

ACCELERATION OF THE CAR = 10 $MS^{-2}$

TIME = 5 SEC.

IT STARTS FROM REST SO ITS INITIAL VELOCITY IS 0 $MS^{-2}$

FINAL VELOCITY = ?

ACCELERATION = $\frac{V-U}{T}$

10 = $\frac{V-0}{5}$

$\frac{10}{1}$ = $\frac{V-0}{5}$

10×5 = 1×V-0

50 = V

DISPLACEMENT = VELOCITY × TIME

SO, 50×5 = 250 M/S

IN IInd CASE

5th SECOND MEANS 5 SEC. ONLY, SO Ist CASE IS SAME AS IInd CASE.

HOPE IT HELPS YOU!

Answer 2

Answer:

Displacement after 5 seconds in 175m.

Explanation:

For constant velocity :

Cobstant velocity u = 10m/s²

Time t = 10s

So, displacement in 10 seconds.

S¹ = ut = 10 × 10 = 100m.

For accelerated journey :

Initial speed u = 10m/s

Acceleration a = 2m/s²

Time t = 5s

Displacement in 5 seconds, S² = ut + ½ at²

==> S² = 10 × 5 + ½ × 2 × 5² = 75m.

Total Displacement in 15 seconds.

S total = 100 + 75 = 175m.