Question

A car of mass 900 Kg is moving with a velocity of 10 m/s. It is brought to rest at 25 m distance by applying the brakes. Calculate the acceleration. Why is it negative? How much force is required to stop the car?​

Answer 1

Answer:

Assumption: the acceleration of this vehicle is constant until it comes to a stop.

Acceleration: $(-2)\; \rm m\cdot s^{-2}$.

The acceleration of this vehicle is negative because the velocity of this vehicle is decreasing over time.

Explanation:

• Let $u$ denote the initial velocity of this vehicle. $u = 10\; \rm m\cdot s^{-1}$.
• Let $v$ denote the final velocity of this vehicle. $v = 0\; \rm m\cdot s^{-1}$ as the vehicle has come to a stop.
• Let $x$ denote the displacement of this vehicle. $x = 25\; \rm m$.
• Let $a$ denote the acceleration of this vehicle. The value of $a\!$ needs to be found.

Assume that the acceleration of this vehicle, $a$, is constant. The SUVAT equations would apply.

The time $t$ required for braking is neither given nor required. Hence, make use of the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to relate $u$, $v$, $a$, and $x\!$.

Rearrange this equation to find an expression for $a$, the acceleration of this vehicle:

$\begin{aligned}a &= \frac{v^{2} - u^{2}}{2\, x} \\ &= \frac{(0\; \rm m\cdot s^{-1})^{2} - (10\; \rm m \cdot s^{-1})^{2}}{2\times (25\; \rm m)} \\ &= \frac{0^{2} - 10^{2}}{2 \times 25}\; \rm m \cdot s^{-2} \\ &= -2\; \rm m\cdot s^{-2}\end{aligned}$.

• When the rate of change of a value is greater than $0$, that value would become larger over time.
• When the rate of change of a value is equal to $0$, that value would be constant.
• When the rate of change of a value is smaller than $0$, that value would become smaller over time.

Acceleration is the rate at which velocity changes over time.

The velocity of the vehicle in this question is getting smaller over time. Hence, the rate of change of the velocity of this vehicle (that is, the acceleration of this vehicle) would be a negative number.