Question

A truck increases its speed from 40 km/hr to 80 km/hr in 10 s . find the acceleration of the truck?​

Answer 1

Answer:

Acceleration = 1.11 m/s²

Explanation:

Initial velocity (u) = 40 km/h

final velocity (v) = 80 km/h

time (t) = 10 sec

To Find - acceleration

Solution -

First lets convert 40 km/h and 60 km/h into m/s

40 X 5/18 = 11.11 m/s

80 X 5/18 = 22.22 m/s

acceleration = v - u / t

                     = 22.22 - 11.11 / 10

                     = 11.11 / 10

                     = 1.11 m/s²

Hope this helped you!!!

Answer 2

The acceleration of the truck is 1.111 m/s².

Explanation

In the question, it's been stated that a truck moves with a velocity of 40 km/hr and suddenly increases to 80 km/hr in 10 seconds. We've been asked to find the acceleration of the truck.

Well, acceleration is the rate of change of velocity with time and it is given by:

$\qquad\qquad\quad \: \sf{\Bigg[ Acceleration = \dfrac{\Delta v}{ \Delta t}}\Bigg]$

Using this formula, the acceleration of the truck can be determined.

• Initial velocity (v₀) = 40 km/hr
• Final velocity = (v) = 80 km/hr
• Time progressed (t) = 10 s

The values of initial velocity and final velocity must be converted to m/s since values in km/hr cannot be used.

In order to convert, their values have to be multiplied with 5/18.

Multiplying the values with 5/18:

$\begin{gathered}\begin{array}{c|c}\sf{ v_{0}= 40 \: kmph}& \sf{v = 80 \: kmph} \\ \\ \twoheadrightarrow\sf{40 \times \dfrac{5}{18} } & \twoheadrightarrow\sf{80 \times \dfrac{5}{18} }\\\\ \twoheadrightarrow\sf{ \dfrac{40 \times 5}{18} }& \twoheadrightarrow\sf{ \dfrac{80 \times 5}{18} }\\\\ \twoheadrightarrow\sf{ \dfrac{200}{18} }& \twoheadrightarrow\sf{ \dfrac{400}{18} } \\ \\ \twoheadrightarrow \sf{ \dfrac{ \cancel2(100)}{ \cancel2(9)} }& \twoheadrightarrow \sf{ \dfrac{ \cancel2(200)}{ \cancel2(9)} }\\ \\ \twoheadrightarrow \sf{ \dfrac{100}{9} }& \twoheadrightarrow \sf{ \dfrac{200}{9} }\\ \\ \twoheadrightarrow \sf{11.11 \: m {s}^{ - 1} }& \twoheadrightarrow \sf{22.22 \: m {s}^{ - 1} }\end{array}\end{gathered}$

All the values are applicable, hence substituting them in the formula:

$\twoheadrightarrow \quad{ \sf{a= \dfrac{\Delta v}{\Delta t} }}$

$\twoheadrightarrow \quad{ \sf{a= \dfrac{ v-v_{0}}{t} }}$

$\twoheadrightarrow \quad{ \sf{a= \dfrac{ 22.22 - 11.11}{10} }}$

$\twoheadrightarrow \quad{ \sf{a= \dfrac{11.11}{10} }}$

$\twoheadrightarrow \quad{ \sf{a= \dfrac{{}^{1111}\!/{}_{100}}{10} }}$

$\twoheadrightarrow \quad{ \sf{a= \dfrac{1111}{100} \times \dfrac{1}{10} }}$

$\twoheadrightarrow \quad{ \sf{a= \dfrac{1111 \times 1}{1000} }}$

$\twoheadrightarrow \quad{ \sf{a= \dfrac{111 1}{1000} }}$

$\twoheadrightarrow \quad \underline{ \boxed{ \bf \red{Acceleration =1.111 \: m/ {s}^{2} }}}$

∴ The acceleration of the truck is 1.111 m/s².