Question

. A car moving with constant acceleration covers the distance between two points 60 m apart

in 6 seconds. Its speed as it passes the second point was 15 m/s.
(a) What is the speed at the

first point?
(b) What is the acceleration?
(c) At what prior distance from the first point was the

car at rest?



Ans:5m/s,
$1 .67m s {}^{2}$
, 7.5m​

Answer 1

Given that, A car is moving with a constant acceleration covers a distance of 60 m ( from A to B ) in 6 seconds.

Also, Its speed as it passes the second point i.e., final velocity, v was found to be 15 m/s

We need to find the speed at the first point, acceleration, and the distance from the first point at which the car was at rest.

Solution :-

We know,

⇒ s = ut + 1/2 at²

Substituting values, we get

⇒ 60 = 6u + 1/2 × a × 6²

⇒ 60 = 6u + 18a

⇒ u + 3a = 10 ...(1)

Also,

⇒ v = u + at

⇒ 15 = u + 6a

⇒ u + 6a = 15 ...(2)

Subtract (1) from (2), we get

⇒ u + 6a - u - 3a = 15 - 10

⇒ 3a = 5

⇒ a = 5 / 3 ...(3)

⇒ a = 1.67 m/s²

Substituting a = 5/3 in (1) from (3), we have

⇒ u + 3×5/3 = 10

⇒ u + 5 = 10

⇒ u = 5 m / s

Now we need to find the distance from the first point at which the car was rest. In this case, the initial velocity becomes final velocity, and initial velocity becomes 0 because the car is at rest.

Using the third Equation of motion,

⇒ 2as = v² - u²

⇒ 2×5/3×s = 5² - 0²

⇒ 10s / 3 = 25

⇒ 10s = 75

⇒ s = 7.5 m

Therefore,

• Acceleration = 1.67 m/s²
• Speed at the first point = 5 m/s
• Distance from the first point at which the car was at rest = 7.5 m