Question

A car starts from rest and accelerates uniformly over a time of 5.1 seconds for a distance of 110 meters.

Answer 1

Given:-

• Initial velocity , u = 0m/s

• Time taken ,t = 5.1s

• Distance ,s = 110m

To Find:-

• Acceleration , a

Solution:-

By using 3rd equation of motion

• s = ut + 1/2at²

Substitute the value we get

→ 110 = 0×5.1 + 1/2×a×5.1²

→ 110 = 0 + 1/2×a× 26.01

→ 110 ×2 = a×26.01

→ 220 = a×26.01

→ a = 220/26.01

→ a = 8.45m/s²

Therefore ,the acceleration of the car is 8.45 m/s²

Answer 2

$\huge \sf \red {\underline {\underline{ GiveN :}}}$

Initial velocity (u) = 0 m/s

Uniform Acceleration

Time interval (t) = 5.21 seconds

Distance travelled (s) = 110 m

$\huge \rm \pink{To \: FinD :}$

Acceleration of the car

$\huge \bf \green{SolutioN :}$

• As we are given that a car starts from rest and accelerates uniformly over time 5.21 seconds, It means car was initially at rest and initial velocity will be 0 m/s. For calculating acceleration, use 2nd equation of motion.

$\sf⇒s = ut + ½at²$

$\sf⇒100 = 0*5.21 + ½ * a * (5.12)²</p><p>$

$\bf⇒100 = 0 + ½ * a * 26.21$

$\sf⇒100 = (26.21 * a)/2</p><p>$

⇒26.21 * a = 110 * 2

⇒26.21 * a = 220

⇒a = 220/26.21

⇒a = 8.3973

$\sf⇒a ≈ 8.4$

∴ Acceleration of the car is 8.4 m/s² (approx.)

_________________________

• Acceleration is Rate of change of velocity.

• It is a vector quantity.

• SI unit of acceleration is m/s².

• Acceleration = Change in velocity/Time

• Instantaneous acceleration is given by dv/dt