Question

A car starts from rest and moves with constant acceleration.
During 4th second of its motion it covers a distance of 21 meters.
The acceleration of the car is
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Answer 1

Answer:

7m/s2

Explanation:

Sn= u+ a/2(2n-1)

21 = 0+ a/2{2(4)-1}

21 = a/2 ×{2×3}

21=3a

a=7m/s2

Answer 2

Given :

• Distance covered in 4th s (S) = 21 m

• nth second (n) = 4 th s

• Initial velocity (u) = 0 m/s

To find :

The Acceleration Produced by the car.

Solution :

Here, the initial velocity is taken as zero because , according to the question , the car starting from rest.

We know the formula for nth second, i.e,

$\boxed{\bf{S_{n} = u + \dfrac{1}{2}a(2n - 1)}}$

Where :

• S = Distance covered
• n = nth second
• a = Acceleration produced
• u = Initial Velocity

Now using the above formula and substituting the values in it, we get :

$:\implies \bf{S_{n} = u + \dfrac{1}{2}a(2n - 1)}$

$:\implies \bf{21 = 0 + \dfrac{1}{2} \times a \times (2 \times 4 - 1)}$

$:\implies \bf{21 = \dfrac{1}{2} \times a \times (2 \times 4 - 1)}$

$:\implies \bf{21 = \dfrac{1}{2} \times a \times (8 - 1)}$

$:\implies \bf{21 = \dfrac{1}{2} \times a \times 7}$

$:\implies \bf{21 = \dfrac{7}{2}a}$

$:\implies \bf{21 \times 2 = 7a}$

$:\implies \bf{42 = 7a}$

$:\implies \bf{\dfrac{42}{7} = a}$

$:\implies \bf{6 = a}$

$\boxed{\therefore \bf{Acceleration\:(a) = 6\:ms^{-2}}}$

Hence, the acceleration produced by the car is 6 m/s².