A car is travelling along a level road. (i) Complete the sentence by putting a cross (X) in the box next to your answer. When the velocity of the car is constant, the force of friction on it is ☐ A zero ☐ B greater than the driving force ☐ C smaller than the driving force ☐ D the same size as the driving force (ii) The car now accelerates in a straight line. Its average acceleration is 12 m/s². Calculate the increase in velocity of the car in 4.0 s. speed = _____________ m/s

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A car is travelling along a level road - Edexcel - GCSE Physics - Question 5 - 2013 - Paper 1

Question 5

A car is travelling along a level road. (i) Complete the sentence by putting a cross (X) in the box next to your answer. When the velocity of the car is constant, t... show full transcript

Worked Solution & Example Answer:A car is travelling along a level road - Edexcel - GCSE Physics - Question 5 - 2013 - Paper 1

Step 1

Complete the sentence by putting a cross (X) in the box next to your answer.

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Answer

When the velocity of the car is constant, the force of friction on it is D: the same size as the driving force. This is because, in a state of constant velocity, the net force acting on an object is zero. Hence, the driving force is balanced by the force of friction.

Step 2

Calculate the increase in velocity of the car in 4.0 s.

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Answer

To calculate the increase in velocity ( v_f - v_i ext{ )}, we use the formula:

v_f = v_i + a imes t

Where:

$v_f$ = final velocity
$v_i$ = initial velocity (which is 0, if starting from rest)
$a$ = acceleration = 12 m/s²
$t$ = time = 4.0 s

Substituting the values, we have:

v_f = 0 + (12 ext{ m/s²}) imes (4.0 ext{ s}) = 48 ext{ m/s}

Thus, the increase in velocity is 48 m/s.

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